# Thread: Introductory Tutorial to Multiple Integration

1. ## Introductory Tutorial to Multiple Integration

As one poster once observed, a small portion of the MHF population answers the majority of the questions. I've found that in regards to multiple integration and vector calculus, that number decreases even more!

I rather enjoy these topics and thought I might write, at least in part, a introductory tutorial on multiple integration and vector calculus. Perhaps no one will use it, but in the case that somebody might get interested who wasnt otherwised exposed to it, that makes it worth it.

Multiple Integration

This is an extremely difficult topic to introduce without the aid of diagrams and to mitigate this problem I'm going to go over a lot of review.

Consider the following problem

Find the area bounded by $y=x$ and $y =x^2$ on the interval $0 \le x \le 1$
When we are first introduced to area problems, we are told to visualize the graph and in this case, it is no different.

Upon drawing the 2 functions we note that $x^2 < x$ from 0 to 1. And since the area is bounded by one function that is higher then the other on the xy axis, we go to our usual formula

$Area = \int_a^b [f(x) - g(x)]dx$ where $f(x) > g(x)$ on the interval a $\to$ b

So for this problem,

$Area = \int_0^1 [ x - x^2 ] dx$

What this actually is however, is a simplified form of a double integral! Note that for the above problem we have the intervals,

$0 \le x \le 1$ and $g(x) \le y \le f(x)$

You will be amazed to know that this is the definition of a double integral!

Definition,

If f(x,y) is continuous on the bounded y-simple domain D given by $a \le x \le b$ and $g(x) \le y \le f(x)$, then

$\iint_D f(x,y)dA = \int_a^b dx \int_{g(x)}^{f(x)} f(x,y)dy$

Similarly,

If f(x,y) is continuous on the bounded x-simple domain D given by $c \le y \le d$ and $g(y) \le x \le f(y)$, then

$\iint_D f(x,y)dA = \int_c^d dy \int_{g(y)}^{f(y)} f(x,y)dx$
So let us re-visit the first problem but this time using multiple integration.

x is bounded by $0 \le x \le 1$ and y is bounded by $x^2 \le y \le x$

Following the form from the above definitions we find that,

$Area = \int_0^1 dx \int_{x^2}^{x} dy$

It's important to note here that we must compute this double integral in a specific order! Since the dy integral deals with functions of x, and our dx integral depends on those functions of x, we cannot evaluate the dx integral first. We must evalute the integral that deals with the most amount of functions in its limits because the remaining integrals depend on them (this will become more clear in triple integration).

So to compute,

$Area = \int_0^1 dx \int_{x^2}^{x} dy = \int_0^1 [x-x^2]dx$

Which is the same result as before!

We should also take this time to note a special property of multiple integration, and that is the indifference of functions.

Consider the following problem

Evaluate the integral: $\int_0^{ 2 \pi } d \theta \int_0^4 dr$
Neither of the integrals in the above problem are dependent on eachother. That is, the $d \theta$ integral doesn't deal in functions of r and the $dr$ integral doesn't deal in functions of theta.

Thereofore it doesn't matter which way we integrate this multiple integration problem

$\int_0^{ 2 \pi } d \theta \int_0^4 dr = \int_0^4 dr \int_0^{ 2 \pi } d \theta = 8 \pi$

At this point I think we're ready to compute a real multiple intregration problem! However, I will lay out what I think are some key steps in evaluating these types of problems

Key Steps

1. Draw the xy domain
2. Select which way you want to integrate (go to the very first definition)
3. Fit your integrals to the general form
4. Compute

Note: Remember that $dA =dxdy$
Consider the following problem

Evaluate $\iint_T (x-3y)dA$ where T is the triangle with the verices $(0,0), (a,0), (0,b)$

So we draw the xy domain and note that this triangle is essentially the line going through $(a,0)$ and $(0,b)$.

At this point we have 2 options. We can either decide to bound our y in terms of x and have x go from one constant to another, or bound our x in terms of y and have y go from one constant to another.

For this problem I think it's easiar to go with the first option. So essentially what we're doing is bounding ourselves by the equation of the line (which we will have to figure out) and the given interval of x (as given in the definition of T). You can think of it as being bounded by the line and the x axis, and shading in all the area underneath that until we hit our interval. In fact, this is the area!

Let's get back to the problem! We need to fit the definition which states,

$I= \int_a^b dx \int_{g(x)}^{f(x)} f(x,y)dy$

So we need to find the equation of our line in terms of x. This is no problem, we will simply fit the curve with $y = mx + b$

Knowing 2 points on the graph $(a,0)$ and $(0,b)$ we find $y = - \frac{b}{a} x + b$

Since this is a triangle in the first quadrent we note that our x is bounded from $0 \le x \le a$

We can now fit this to the definition and find that,

$I = \int_0^a dx \int_0^{ - \frac{b}{a} x + b } (x-3y) dy$

This is a great problem because it introduces us to a wide variety of things! I will break it up so that we can see step by step where it all comes from, but soon you will be able to do this all in your head without going through these mini steps

$\int_0^a dx \int_0^{ - \frac{b}{a} x + b } (x-3y) dy = \int_0^a dx \int_0^{ - \frac{b}{a} x + b } (x)dy + \int_0^a dx \int_0^{ - \frac{b}{a} x + b } (-3y)dy$

The same properties that hold for single integration hold for double integration, so we can take constants out!

$I = \int_0^a dx \int_0^{ - \frac{b}{a} x + b } xdy -3 \int_0^a dx \int_0^{ - \frac{b}{a} x + b } ydy$

Let's look at the first part,

Note that: $I = I_1 + I_2$

$I_1 = \int_0^a dx \int_0^{ - \frac{b}{a} x + b } xdy$

Remember when I said we can treat functions that the integral doesn't depend on as constants? Well in this case we have x multiplied by dy, but since we only integrate with respect to y in the dy integral, we can move the x to its appropriate place!

$\int_0^a dx \int_0^{ - \frac{b}{a} x + b } xdy = \int_0^a xdx \int_0^{ - \frac{b}{a} x + b } dy$

Now, our dy integral deals in functions of x, so we must compute this first!

$\int_0^a x (- \frac{b}{a} x + b) dx = \int_0^a [bx - \frac{b}{a} x^2 ] dx = [ \frac{b}{2} x^2 - \frac{b}{3a} x^3]_0^a = \frac{b}{2} a^2 - \frac{b}{3} a^2$

Now for the second part of the integral we have,

$I_2 = -3 \int_0^a dx \int_0^{ - \frac{b}{a} x + b } ydy$

$= -3 \int_0^a \frac{[b - \frac{b}{a} x]^2}{2} dx$

$= [-\frac{3}{2} b^2 x + \frac{3}{2} \frac{b^2 x^2}{a} - \frac{1}{2} \frac{b^2 x^3}{a^2}]_0^a = -\frac{3}{2} b^2 a + \frac{3}{2} b^2 a - \frac{1}{2} b^2 a$

Summing the first part and second part of this integral we get,

$I= I_1 + I_2 = \frac{b}{2} a^2 - \frac{b}{3} a^2 -\frac{3}{2} b^2 a + \frac{3}{2} b^2 a - \frac{1}{2} b^2 a = \frac{a^2 b}{6} - \frac{ab^2}{2}$

You solved a multiple integration problem! Was that as hard is you thought? Definately not! It's fairly easy once you get the hang of it.

Now there are some tricks to multiple intregation, one of which is a change of bounds.

Going back to the definition of a double integral we can either bound of y in terms of x or our x in terms of y.

When we are given a problem such that

$\int_a^b dx \int_{g(x)}^{f(x)} f(x,y)dy$

and we change it to

$\int_c^d dy \int_{g(y)}^{f(y)} f(x,y)dx$

it is known as a change of bounds.
We do this when the integral, in its present form, is relatively hard to compute.

Consider the following problem

Evaluate the following integral: $\iint_R \frac{x}{y} e^y dA$ where R is the region $0 \le x \le 1$ and $x^2 \le y \le x$
We've seen this xy domain in a previous problem, but in this case our $f(x,y) = \frac{x}{y} e^y$ is relatively messy.

If we set up the integral as given we get,

$\int_0^1 xdx \int_{x^2}^x \frac{e^y}{y} dy$

To put it simply, there is no good way to integrate $\int_{x^2}^x \frac{e^y}{y} dy$. While there are tricks, they are hard and we don't know them at this stage!

In this case we can employ a change of bounds! We can turn our functions of $y=f(x)$ into functions such that $h(y) = x$ and go side to side (as opposed to top to bottem) and have constant bounds for our y domain.

Let's look at our functions,

$x^2 \le y \le x$ leads to $y = x^2 \le y \le y= x$

$y =x^2 \to x = \sqrt{y}$ and $y=x \to x=y$

Remember that our functions intersect at x=0 and x=1, but they also intersect at y=0 and y=1!

So our constant y bounds are $0 \le y \le 1$

And on that interval, $y \le \sqrt{y}$

Thus,

$\int_0^1 xdx \int_{x^2}^x \frac{e^y}{y} dy = \int_0^1 \frac{e^y}{y} dy \int_y^{ \sqrt{y}} xdx$

I want to take a moment right here and point at our bounds. When I changed them, it may have seemed simple and all we're doing is reversing them (i.e. $y=x^2 \to x= \sqrt{y}$ ) but this is not the case. If we didn't have to idenitfy any further we would think,

$\int_0^1 xdx \int_{y=x^2}^{y=x} \frac{e^y}{y} dy = \int_0^1 \frac{e^y}{y} dy \int_{\sqrt{y}}^{ y}xdx$

Which is not true, so it's important that we draw the xy domain so we can identify what our bounds actually are depending on the interval!

Back to the problem,

$\int_0^1 \frac{e^y}{y} dy \int_y^{ \sqrt{y}} xdx = \frac{1}{2} \int_0^1 (1-y)e^y dy$

Let $U = 1-y$ and $dV = e^y dy$

Such that $du = -dy$ and $V =e^y$

$\frac{1}{2} \int_0^1 (1-y)e^y dy = \frac{1}{2} [ [(1-y)e^y]_0^1 + \int_0^1 e^y dy] = -\frac{1}{2} + \frac{1}{2} (e-1) = \frac{e}{2} - 1$

Bellow are some homework questions for multiple integration.

1: $\iint_S (sinx + cosy)dA$ where S is the square $0 \le x \le \frac{ \pi }{2}$ and $0 \le y \le \frac{ \pi }{2}$

Spoiler:
$\iint_S (sinx + cosy)dA$

$= \int_0^{ \frac{ \pi}{2}} dx \int_0^{ \frac{ \pi}{2}} (sinx + cosy)dy$

$= \int_0^{ \frac{ \pi}{2}} dx [ysinx + siny]_0^{\frac{ \pi}{2}}$

$= \int_0^{ \frac{ \pi}{2}} [ \frac{ \pi}{2} sinx + 1] dx$

$= \pi$

2: $\iint_D lnx dA$ where D is the finate region in the first quadrant bounded by the line $2x+2y = 5$ and the hyperbola $xy=1$

Spoiler:

$\iint_D lnx dA$

Intersection of the line $2x+2y = 5$ and the hyperbola $xy=1$ happens when

$2x^2 -5x + 2 = (2x-1)(x-2)=0$

The intersections are at $x = \frac{1}{2}$ and $x = 2$

$\iint_D lnx dA$

$\int_{ \frac{1}{2}}^2 lnxdx \int_{\frac{1}{x}}^{ \frac{5}{2} - x } dy$

$= \int_{ \frac{1}{2}}^2 lnxdx ( \frac{5}{2} - x - \frac{1}{x} )dx$

$= \int_{ \frac{1}{2}}^2 lnxdx ( \frac{5}{2} - x)dx - [\frac{1}{2} (lnx)^2 ]_{\frac{1}{2}}^2$

Let $U =lnx$ and $dV=( \frac{5}{2} - x ) dx$

Such that $dU = \frac{dx}{x}$ and $V = \frac{5}{2}x - \frac{x^2}{2}$

$= - \frac{1}{2} [ (ln2)^2 - (ln \frac{1}{2} )^2] + [( \frac{5}{2} x - \frac{x^2}{2})lnx]_{ \frac{1}{2} }^2 - \int_{ \frac{1}{2} }^2 ( \frac{5}{2} - \frac{x}{2} ) dx$

$= (5-2)ln2 ( \frac{5}{4} - \frac{1}{8} )ln \frac{1}{2} - \frac{15}{4} + \frac{15}{16}$

$= \frac{33}{8}ln2 - \frac{45}{16}$

3: $\iint_T \frac{xy}{1 + x^4 } dA$ where T is the triangle with the vertices (0,0), (1,0), (0,1)

Spoiler:

$\iint_T \frac{xy}{1 + x^4 } dA$

$= \int_0^1 \frac{ x}{1 + x^4 } dx \int_0^x ydy$

$= \frac{1}{2} \int_0^1 \frac{x^3}{1+x^4}dx$

$= \frac{1}{8} ln(1 + x^4)_0^1$

$= \frac{ln2}{8}$

4: $Evaluate: \int_0^{ \frac{ \pi }{2} } dy \int_y^{ \frac{ \pi }{2}} \frac{ sinx}{x} dx$

Spoiler:

$\int_0^{ \frac{ \pi }{2} } dy \int_y^{ \frac{ \pi }{2}} \frac{ sinx}{x} dx$

$= \iint_R \frac{sinx}{x} dA$

$= \int_0^{ \frac{ \pi }{2}} dx \int_0^x dy$

$= \int_0^{ \frac{ \pi}{2}} sinx dx$

$= 1$

2. Double Integral in Polar Coordinates

Up until now we have dealt with double integrals dealing with bounds in terms of x and y, known as cartesian coordinates. But what happens if this becomes relatively hard to do?

Consider the example of a circle, would you compute the area of a circle (if we didnt know that $A = \pi r^2$) in cartesian coordinates? Probably not,

Using single integration we would probably find the area of the half hemisphere and multiply it by 2, that is

$\frac{A}{2} = \int_{-r}^{r} \sqrt{ r^2 - x^2} dx$

Assuming we didnt know the area that this integral represented, we would use the subsitution $x = rsin \theta$

However that's single integral thinking! And we now have double integrals by which to calculate these sorts of things. So how do we do that?
In polar co-ordinates we let

$x = r cos \theta$

$y = rsin \theta$

$r^2 = x^2 + y^2$

$dA = rdr d \theta$
We tend to use cylindrical co-ordinates when there is a cylinder or sphere involved in the calculation.

Noting that r denotes radius, we can easily compute the area of a circle.

$A = \int_0^{ 2 \pi } d \theta \int_0^a r dr = \pi a^2$

If the above confuses you, don't worry! Let's go through it again.

Well...the first thing is to visualize. When we have a circle centered at the origin we can make a line from the origin (0,0) to any point on the circle of radius a. Now sweep that line, from (0,0) to a, all 360 degrees. You can even take your finger and do this. Notice how we've covered the entire circle! That's what double integration is, we are essentially sweeping over the area and by integrating with respect to bounded functions, we can obtain that area.

Consider the following problem

Evaluate: $\iint_D \frac{dA}{ \sqrt{x^2 + y^2}}$ where D is the circle of radius a.
In carestian coordinates the above could be quite messy. I mean, we're looking at a lot of substitution! So instead, lets sub in polar coordinates

That means,

$\iint_D \frac{dA}{ \sqrt{x^2 + y^2}} = \iint_Q \frac{rdr d \theta }{ \sqrt{r^2}} = \iint_Q dr d \theta$ where Q is the domain of a circle of radius a.

Doesn't that look a lot more simple?! Clearly we find that

$\iint_D \frac{dA}{ \sqrt{x^2 + y^2}} = \iint_Q dr d \theta = \int_0^{ 2 \pi } d \theta \int_0^a dr = 2 \pi a$

That was a simple example, and these problems can become much more complex. So let us investigate a problem where our r does not cancel out so nicely. We will find that sketching our domain is crucial!

Consider the following problem

Evaluate: $\iint_s (x+y) dA$ where S is the region bounded by the line $y = \sqrt{3}x$ and the circle $x^2 + y^2 = a^2$ in the first quadrant
Woha, this problem is introducing 2 equations on top of an integral!? Don't fret because while this may seem to be an overload of information, it actually isn't anything new.

First, we have to find our finite region. Since we are bounded by 2 equations we should draw these and figure out what the shape looks like.

$x^2 +y^2 = a^2$ is nothing more then a circle of radius a. So draw that on paper.

The line $y = \sqrt{3} x$ is simply a line that goes through the origin, that will eventually hit the circle. Notice how how the region bounded by our 2 equations (the circle and the line) is the entire area of the circle UP UNTIL the line? We don't want the area overtop of this line! We want the area underneath but contained within the circle.

But how are we going to compute this? Should we keep in cartesian coordinates or polar? Since we have a circle as part of our domain it might be best to transform to polar coordinates, so for this problem, we will use polar.

So lets find our domain for both theta and r.

Unfortunately, our theta domain is no longer $2 \pi$! Recall that for our earlier problems our theta domain was 360 degrees, but that's because we wanted to go the entire way around the circle!

In this case, we only want to go sweep from the x-axis to our line. Well, our line has an angle of $\frac{ \pi }{3}$! We know this because of special triangles. If you notice the rise over run of one of the special triangles is $\sqrt{3}$ so we can use that result: Special right triangles - Wikipedia, the free encyclopedia

We now know our theta bounds! $0 \le \theta \le \frac{ \pi }{3}$ in case you forget, our angle is measured from the x-axis in a counter clockwize manner.

But what about our bounds for r? Well...We go from the origin to the radius of the circle, so our r bounds must be from 0 to a.

Thus,
$\iint_s (x+y) dA$

$= \int_0^{ \frac{ \pi }{3}} d \theta \int_0^a (rcos \theta + r sin \theta) r dr$

$= \int_0^{ \frac{ \pi }{3}} (cos \theta + sin \theta )d \theta \int_0^a r^2 dr$

$= \frac{ a^3}{3} [ sin \theta - cos \theta ]_0^{ \frac{ \pi}{3}}$

$= [ ( \frac{ \sqrt{3} }{2} - \frac{1}{2} ) - (-1)] \frac{a^3}{3}$

$= \frac{ ( \sqrt{3} + 1) a^3 }{6}$

Note how we sub in $x = r cos \theta$ and $y = r sin \theta$

So far we have dealt with static domains, and what I mean by that is r does not change with respect to theta. In the case of the above problem we are always drawing a line from the origin to the radius of the circle and sweeping that over the region.

But what if we have a shape where r is a function of theta? How does this change the problem?

Consider the following problem

Evaluate: $\iint_T (x^2 + y^2)dA$ where T is the region bounded by the triangle with the verticies (0,0), (1,0), (1,1)
If we draw the domain we can clearly see this is a triangle with the hypotenuse being bounded by the line $y = x$ this means our angle theta is $\frac{ \pi }{4}$ by special triangles.

So right away, we have our theta bounds

$0 \le \theta \le \frac{ \pi }{4}$

But what about r? If you think about it, the length of the hypotenuse will change with the inclination, so we need to find a value of r such that is relates to theta!

Draw another triangle and label the angle theta, the bottem of the triangle 1, and the hypotenuse r. Now take,

$cos \theta = \frac{1}{r}$

$r = \frac{1}{cos \theta } = sec \theta$

The reason why the bottem of the triangle is 1 is because this will never change, no matter what our x value is 1 and it's only the length of our hypotenuse that changes.

We can now compute this problem

$\iint_T (x^2 + y^2)dA = \int_0^{ \frac{ \pi }{4} } d \theta \int_0^{ sec \theta } r^3 dr$

$= \frac{1}{4} \int_0^{ \frac{ \pi }{4}} sec^4 \theta d \theta$

$= \frac{1}{4} \int_0^{ \frac{ \pi }{4}}(1 + tan^2 \theta )sec^2 \theta d \theta$

Let $u = tan \theta$ and $du= sec^2 \theta d \theta$

$= \frac{1}{4} \int_0^1 (1+u^2) du$

$= \frac{1}{3}$

Keeping with the idea of shifting our domain, what about a circle that isn't centered at the origin? Thus far, we have computed problems using polar co-ordinates where r is from the origin to the radius of the circle. Clearly, this won't work if the circle is centered somewhere else other then the origin!

Consider the following problem

Find the volume bounded by the cylinder $x^2 + y^2 = ax$ and the hemisphere $z = \sqrt{ a^2 - x^2 - y^2 }$
First thing to note is that our cylinder is no longer centered at the origin.

$x^2 + y^2 = ax$

$(x^2-ax) + y^2 = 0$

$(x- \frac{a}{2})^2 - \frac{a^2}{4} + y^2 = 0$

$(x- \frac{a}{2})^2 + y^2 = \frac{a^2}{4}$

So we can see that our circle is centered on the x-axis at the coordinate

$( \frac{a}{2} , 0 )$

Not a problem, let's find our r domain

$x^2 + y^2 = ax$

Remember that $x = r cos \theta$

So

$x^2 + y^2 = r^2 = ax = a rcos \theta$

$r = a cos \theta$

So our r bounds are $0 \le r \le a cos \theta$

Our integral becomes,

$Volume = 4 \iint_D \sqrt{ a^2 - x^2 -y^2} dA$

Where D is the domain of the circle in the first quadrant.

Take a moment to look at the picture. We want the volume between the hemisphere and the cylinder. Which is to say, we want the sphere projected over the domain of the cylinder.

I know I know, this isn't making a whole lot of sense but think of it like this,

$\iint_D \sqrt{ a^2 - x^2 -y^2} dA$

If D is the domain of the hemisphere , this represents the volume of a hemisphere inside a hemisphere. However, if D represents a circle (or a cylinder) this integral represents the volume of a hemisphere INSIDE the domain of a cylinder.

The 4 in

$Volume = 4 \iint_D \sqrt{ a^2 - x^2 -y^2} dA$

comes from our bounds, we are going to bound ourselves to the first quadrant and multiply by 4 to get the entire hemisphere.

So to compute,

$Volume = 4 \iint_D \sqrt{ a^2 - x^2 -y^2} dA$

$= 4 \int_0^{ \frac{\pi}{2}} d \theta \int_0^{acos \theta} \sqrt{a^2-r^2} rdr$

Note that our theta is bounded by 0 and $\frac{ \pi }{2}$ because our value for r, that being $r = a cos \theta$ is valid for the region $0 \le \theta \le \frac{ \pi }{2}$

If you notice,

$4 \int_0^{ \frac{ \pi }{2} } = \int_0^{ 2 \pi } = 2 \pi$

$= 4 \int_0^{ \frac{\pi}{2}} d \theta \int_0^{acos \theta} \sqrt{a^2-r^2} rdr$

Let $u = a^2-r^2$ and $du = -2rdr$

$= 2 \int_0^{ \frac{\pi}{2}} d \theta \int_{a^2 sin^2 \theta}^{a^2} u^{ \frac{1}{2}} du$

$= \frac{4}{3} \int_0^{ \frac{ \pi}{2}} d \theta [ u^{ \frac{3}{2}}]_{a^2 sin^2 \theta}^{a^2}$

$= \frac{4}{3} a^3 \int_0^{ \frac{\pi}{2}} (1- sin^3 \theta )d \theta$

$= \frac{4}{3} a^3 [ \frac{ \pi }{2} - \int_0^{ \frac{\pi}{2}} sin \theta (1- cos^2 \theta )d \theta ]$

Let $v = cos \theta$ and $dv=-sin \theta d \theta$

$= \frac{ 2 \pi a^3 }{3} - \frac{4a^3}{3} \int_)^1 (1-v^2)dv$

$= \frac{ 2 \pi a^3 }{3} - \frac{4a^3}{3} (v- \frac{v^3}{3})_0^1$

$=\frac{ 2 \pi a^3 }{3} - \frac{ 8 a^3}{9}$

$= \frac{2}{9} a^3 (3 \pi - 4)$

That's it for the instruction on basic polar coordinates dealing with double integrals. I will leave some homework problems for you!
1: Find the area between the paraboloids $z = x^2 + y^2$ and $3z = 4 -x^2 -y^2$
Spoiler:

$z = x^2 + y^2$ and $3z = 4 -x^2 -y^2$ intersect where

$3(x^2 +y^2) = 4 - (x^2 +y^2) \to x^2 + y^2 = 1$

This represents our xy domain,

$V=\iint_{x^2 +y^2 \le 1 } [ \frac{ 4 - x^2 -y^2 }{3} -(x^2+y^2)] dA$

Note that the integrand comes from subtracting the smaller function of Z from the larger function of Z, the same as we do for single integration in our calculation of area.

$= \int_0^{ 2 \pi } d \theta \int_0^1 [ \frac{ 4 -r^2}{3} -r^2] rdr$

$= \frac{8 \pi }{3} \int_0^1 (r-r^3) dr$

$= \frac{2 \pi }{3}$

2: Evaluate: $\iint_S x dA$ where S is the area that lies outside the circle $x^2 + y^2 = 2$ and the line $y = x$
Spoiler:

This is a combination of past problems. Drawing the domain we see that we want 2 components, one of the left side of our circle and one on the right. We will compute for 1 and multiply by 2 to get the total volume.

In this case we have a minimum r that is no 0! If you draw a line down from the line where it hits the circle, all the way to the x-axis, we can see that we dont want our r anywhere from 0 to this point.

Thus,

$sec \theta \le r \le \sqrt{2}$

Where $sec \theta$ comes from our evluation of a past problem.

$\iint_S x dA$

$= 2 \int_0^{ \frac{ \pi }{4}} d \theta \int_{ sec \theta }^{ \sqrt{2} } r cos \theta dr$

$= \frac{2}{3} \int_0^{ \frac{ \pi }{4}} cos \theta ( 2 \sqrt{2} - sec^3 \theta) d \theta$

$= \frac{ 4 \sqrt{2}}{3} [ sin \theta ]_0^{ \frac{ \pi }{4} } - \frac{2}{3} [ tan \theta ]_0^{ \frac{ \pi }{4} }$

$= \frac{ 4}{3} - \frac{2}{3}$

$= \frac{2}{3}$

3: Evaluate: $\iint_{x^2 +y^2 \le 1 } ln(x^2 + y^2) dA$

Spoiler:

$= \int_0^{ 2 \pi } d \theta \int_0^1 (lnr^2)rdr$

$= 4 \pi \int_0^1 rlnr dr$

Let $U = lnr$ and $dV = rdr$

Such that $du= \frac{dr}{r}$ and $V = \frac{r^2}{2}$

$= 4 \pi [ (\frac{r^2}{2} lnr)_0^1 - \frac{1}{2} \int_0^1 rdr]$

$= 4 \pi [ 0-0- \frac{1}{4}]$

$= - \pi$

Note that the integral is improper, but converges absolutely $\lim_{ r \to 0^+ } r^2 lnr = 0$

3. Change of variables in double integrals

Now that we've talked about polar and cartesian co-ordinates, i want to bring to your attention what is known as a change of variables.

Say we have,

$\iint_D f(x,y) dA$

But at first glance, this looks very tricky. We can change it such that,

$x = g(u,v)$ and $y = h(u,v)$

When we alter our variables in such a way it is known as changing with the Jacobian, or the Jacobian transformation.

In the Jacobian transformation,

$dA = \left|\frac{ \partial (x,y) }{ \partial (u,v) } \right| dudv = \left| \begin{matrix} { \frac { \partial x}{u}} & { \frac { \partial x}{ v}} \\ { \frac { \partial y}{u}} & { \frac { \partial y}{v}} \end{matrix} \right | dudv$

The definition is

$\iint_D f(x,y) dA = \iint_P f( g(u,v), h(u,v) ) \left|\frac{ \partial (x,y) }{ \partial (u,v) } \right| dudv$
Consider the following problem

Find the volume of the ellipsoid $\frac{z^2}{c^2} + \frac{ x^2}{a^2} + \frac{ y^2}{b^2}$
Writing this out we get,

$V = 2c \iint_D \sqrt{ 1 - \frac{ x^2}{a^2} - \frac{ y^2}{b^2} } dA$

Yikes! The above can't be dealt with in terms of normal methods very easily...so we can use the Jacobian here

Let $x = au$ and $y = bv$

Then,

$dxdy = \left|\frac{ \partial (x,y) }{ \partial (u,v) } \right| dudv = \left| \begin{matrix} { \frac { \partial x}{u}} & { \frac { \partial x}{ v}} \\ { \frac { \partial y}{u}} & { \frac { \partial y}{v}} \end{matrix} \right | dudv$

$= \left| \begin{matrix} { a} & { 0} \\ { 0} & { b} \end{matrix} \right | dudv = ab dudv$

Therefore,

$V = 2c \iint_D \sqrt{ 1 - \frac{ x^2}{a^2} - \frac{ y^2}{b^2} } dA$

$= 2abc \iint_Q \sqrt{ 1 - u^2 - v^2} dudv$

Where Q is the the disk $u^2 + v^2 \le 1$

Q comes from letting $z = 0$ because we are only in the xy plane and z is always 0

Recall that after the transformation $z = \sqrt{ 1 - u^2 - v^2} \to u^2 + v^2 \le 1$

We can actually note here that

$\iint_Q \sqrt{ 1 - u^2 - v^2} dudv$

is simply the volume of a ball of radius 1 divided by 2 (because we are only in the upper hemisphere as defined by our region Q).

So,

$= 2abc \iint_Q \sqrt{ 1 - u^2 - v^2} dudv$

$= 2abc \frac{ 4\pi 1^2 }{3 * 2}$

$= abc \frac{ 4 \pi }{3}$

That's essentially all there is to the Jacobian in double integrals. The most important aspect is knowing when to use the transformation and what part of the equation we should actually transform.

The following homework problems will give you a good sense of when to use the Jacobian.

1: Evaluate: $\iint_P (x^2 +y^2)dxdy$ where P is the paralleligram bounded by $x+y=1$, $x+y = 2$, $3x + 4y = 5$, and $3x+4y = 6$

Spoiler:

Let $u = x + y$ and $v = 3x + 4y$

This corresponds to the following diagram

$\left|\frac{ \partial (x,y) }{ \partial (u,v) } \right| dudv$

$= \left| \begin{matrix} { 4} & { -1} \\ { -3} & { 1} \end{matrix} \right | dudv = 1 dudv$

Also note,

$x^2 + y^2 = (4u-v)^2 + (v-3u)^2 = 25u^2 - 14uv +2v^2$

Thus,

$\iint_P (x^2 +y^2)dxdy$

$= \iint_S (25u^2 - 14uv +2v^2) dudv$

$= \int_1^2 du \int_5^6 (25u^2 - 14uv +2v^2) dv$

$= \frac{7}{2}$

2: Evaluate: $\iint_T e^{ \frac{ (y-x) }{ (y+x)}}$ where T is the triangle bounded by the line $x+y = 1$ and $x,y \ge 0$

Spoiler:

Let $u = y-x$ and $v = y+x$

Then this becomes the diagram,

T` is bounded by $u = v$ and $u = - v$

To find the Jacobian, find x and y only in terms of u and v. We will get,

$x = \frac{v}{2} - \frac{u}{2}$

$y = \frac{v}{2} + \frac{u}{2}$

$\left|\frac{ \partial (x,y) }{ \partial (u,v) } \right| dudv$

$= \left| \begin{matrix} { - \frac{1}{2} } & { \frac{1}{2} } \\ { \frac{1}{2} } & { \frac{1}{2} } \end{matrix} \right | dudv$

$= -\frac{1}{2} dudv$

Thus,

$\iint_T e^{ \frac{ (y-x) }{ (y+x)}}$

$= - \frac{1}{2} \int_0^1 dv \int_{v}^{-v} e^{ \frac{u}{v}} du$

$= \frac{1}{2} \int_0^1 dv(ve^{ \frac{ u}{v}})_{-v}^v$

$= \frac{1}{2} (e - e^{-1} ) \int_0^1 vdv$

$= \frac{ e -e^{-1}}{4}$

4. Triple Integration

Now that we've been exposed to multiple integration problems it becomes a matter of exercise to learn triple integration.

Triple integration is defined as

$\iiint_B f(x,y,z) dV = \iiint_B f(x,y,z) dxdydz$
We go through our same steps as we did in double integration, except we have one more integral.

In general,

$\iiint_B f(x,y,z) dV = \int_a^b dx \int_{h_1(x)}^{h_2(x)} dy \int_{g_1(x,y)}^{g_2(x,y)} f(x,y,z) dz$
There are some special properties inherent with triple integration. Namely, we can evaluate triple integrals by inspection, using symmetry and using known volumes.

Consider,

Volume of D $= \iiint_D dV$

Which means the triple integral of only a domain (no function), gives the volume of that domain.

Similarly,

If a domain is symmetric about a coordinate plane (for example, symmetrical about x=0 and y=0) then

$\iiint_D x = \iiint_D y = \iiint_D xy = 0$

Consider the following problem

Evaluate $\iiint_R (1+2x-3y)dV$, over the box $-a \le x \le a$, $-b \le y \le b$, and $-c \le z \le c$
$\iiint_R (1+2x-3y)dV$

We note that $\iiint_R 2x dV$ and $\iiint_R -3y dV$ are both equal to 0 because of symmetry.

If you can't see this, evaluate both of the integrals and note how the result is 0.

Our integral reduces to

$\iiint_R dV$

$=Volume of the box$

$= 8abc$

So you can see of symmetry and inspection reduces our work load, if detected early on.

Consider the following problem
$\iiint_R (xy + z^2)dV$ over the set $0 \le z \le 1 - |x| - |y|$
Here, we will first draw the diagram noting that the given region is the pyramid, 1 quarter of which lies in the first octant.

Since we are confining ourselves to the first octant our equation becomes

$x + y + z = 1$ which means by symmetry, $xy$ over R is equal to 0.

Thus,

$\iiint_R (xy + z^2)dV$

$= \iiint_R z^2 dV$

$= 4 \int_0^1 z^2 dz \int_0^{1-z} dy \int_0^{1-z-y} dx$

The above bounds of integration come from letting our equation equal to 0 in the respective planes. So, in the dydz plane $x = 0 \to 1-z$ which is clearly greater then 0, so this is our top function.

The 4, once again, comes from bounding ourselves in the first quadrent. There are 4 quadrents so we multiply by 4

$= 4 \int_0^1 z^2 dz \int_0^{1-z} (1-z-y)dy$

$= 4 \int_0^1 z^2 [(1-z)^2 - \frac{1}{2} (1-z)^2] dz$

$= 2 \int_0^1 (z^2 - 2 z^3 + z^4)dz$

$= \frac{1}{15}$

I will leave you with only 1 homework question this time! I know, i'm too nice

1: $\iiint_R sin( \pi y^3 ) dV$ over the pyramid with the vertices (0,0,0), (0,1,0), (1,1,0), (1,1,1), and (0,1,1)

Spoiler:

It's very important to draw the domain here. Since we can integrate this in any way we want, we should want to pick integrating dy last. The reason for this is the y cubed in the original function.

To integrate that into another function...well, that's just messy. In generl, if our function contains some high order variable and we can pick which way we want to integrate, then we integrate with respect to that variable last.

This is where the diagram becomes important,

Note how in the xy domain, x=y and in the zy domain z=y. We can tell this because we have points in these domains that are straight lines through the point (1,1) (or (0,1,1) in the case of the yz domain).

This means, our bottem function for both our dz and dx integrals is 0 and our top functions are y!

Thus,

$\iiint_R sin( \pi y^3 ) dV$

$= \int_0^1 sin( \pi y^3 ) dy \int_0^y dz \int_0^y dx$

$= \int_0^1 sin( \pi y^3 )dy$

$= - \frac{cos( \pi y^3 )}{3 \pi } |_0^1$

$= \frac{2}{3 \pi}$

Polar Coordinates in Triple Integrals

In triple integrals, polar coordinates don't change all that much.

For polar coordinates with triple integrals

$x = r cos \theta$

$y = r sin \theta$

$z=z$

$dV = r dr d \theta dz$
Consider the following problem

Find the volume between the paraboloids $z = 10 - x^2 - y^2$ and $z =2(x^2 + y^2 - 1)$

First thing to note is that even if we cant tell what the diagram looks like (though we have one drawn here) we can look for recognizable patterns. Notice how both paraboloids contain either $- x^2 - y^2$ or $x^2 + y^2$

Well we know what $y = x^2$ looks like, but in this case we have it in the zxy domain.

So basically we can tell right off the hop we have one parabloid opening downward and one uponing upward.

To find our xy domain we will equate these paraboloids (after transforming to polar coordinates) to get

$2(r^2 - 1) = 10 -r^2$ which intersect when $r = 2$

Thus, the volume between the surfaces is

$\int_0^{ 2 \pi } d \theta \int_0^2 r dr \int_{ 2(r^2 - 1) }^{ 10 -r^2} dz$

$= 2 \pi \int_0^2 (12r-3r^3) dr$

$= 24 \pi$

Now lets compute another example that requires a knowledge of cone like shapes (cones are very important in triple integration, they come up a lot!)

Consider the following problem

Find the volume above the surface $z = (x^2 + y^2)^{ \frac{1}{4} }$ and inside the sphere $x^2 + y^2 + z^2 = 2$
Transforming to polar we get

$z = (r^2)^{ \frac{1}{4} } = \sqrt{r}$

and

$z = \sqrt{2-r^2}$

To find the intersection we equate them to get

$r^2 + r - 2 = 0$ which has the posative root $r = 1$

Drawing the domain,

Thus,

$V = \int_0^{ 2 \pi} d \theta \int_0^1 r dr \int_{ \sqrt{r}}^{ \sqrt{2-r^2} } dz$

$= \int_0^{ 2 \pi} d \theta \int_0^1 [ \sqrt{2-r^2} - \sqrt{r} ]r dr$

$= 2 \pi [ \int_0^1 r \sqrt{2-r^2} dr - \frac{2}{5} ]$

Let $u = 2 - r^2$ and $du = -2rdr$

$= \pi \int_0^2 u^{ \frac{1}{2} } du - \frac{4 \pi }{5}$

$= \frac{ 4 \sqrt{2} \pi }{3} - \frac{22 \pi }{15}$

Here are some homework questions for you to try

1: Find the volume inside the paraboloid $z = x^2 + y^2$ and inside the sphere $x^2 +y^2 + z^2 = 12$

Spoiler:

Transforming to polar we get

$z = r^2$ and $r^2 + z^2 = 12$

Equating these yields $r^4 + r^2 - 12 =0$ which has the root $r = \sqrt{3}$

Thus, the required volume is

$V = \int_0^{ 2 \pi } d \theta \int_0^{ \sqrt{3} } ( \sqrt{12 - r^2} - r^2)rdr$

$= 2 \pi \int_0^{ \sqrt{3} } r \sqrt{12 - r^2} dr - \frac{ 9 \pi }{2}$

Let $u = 12 - r^2$ and $du = -2rdr$

$= \pi \int_9^{12} u^{ \frac{1}{2}} du - \frac{ 9 \pi }{2}$

$= 16 \sqrt{3} \pi - \frac{45 \pi }{2}$

2: Find the volume above the xy plane, under the paraboloid $z= 1 - x^2 - y^2$ and in the wedge $-x \le y \le \sqrt{3} x$

Spoiler:

Transforming to polar

$z = 1-r^2$

Note that,

$-x \le y \le \sqrt{3} x$

Correponds to the angles

$- \frac{ \pi }{4} \le \theta \le \frac{ \pi }{3}$

Thus,

$V = \int_{ \frac{ \pi }{4} }^{ \frac{ \pi }{3} } d \theta \int_0^1 r dr \int_0^{ 1-r^2} dz$

$= \int_{ \frac{ \pi }{4} }^{ \frac{ \pi }{3} } d \theta \int_0^1 (1-r^2) r dr$

$= \frac{ 7 \pi }{12} ( \frac{1}{2} - \frac{1}{4} )$

$= \frac{ 7 \pi }{48}$

3: Evaluate $\iiint_R ( x^2 + y^2 + z^2 ) dV$ where R is the cylinder $0 \le x^2 + y^2 \le a^2$, $0 \le z \le h$

Spoiler:

$\iiint_R ( x^2 + y^2 + z^2 ) dV$

$= \int_0^{ 2 \pi } d \theta \int_0^{ a } rdr \int_0^h (r^2 + z^2) dz$

$= 2 \pi \int_0^a ( r^3h + \frac{1}{3} rh^3 )dr$

$= 2 \pi [ \frac{ a^4h}{4} + \frac{a^2 h^3}{6} ]$

Triple Integrals and Spherical coordiates

There is another co-ordinate system known as spherical co-ordinates. However, it's rather hard to explain over an internet forum so I will default to Paul's notes; they are very good.

Well...that's it for multiple integration. I will add vector calculus when I have more time!

5. ## Re: Introductory Tutorial to Multiple Integration

This is a great information to everyone.