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

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

Upon drawing the 2 functions we note that \(\displaystyle x^2 < x \) 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

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

So for this problem,

\(\displaystyle Area = \int_0^1 [ x - x^2 ] dx \)

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

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

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

__Definition__,

So let us re-visit the first problem but this time using multiple integration.If f(x,y) is continuous on the bounded y-simple domain D given by \(\displaystyle a \le x \le b \) and \(\displaystyle g(x) \le y \le f(x) \), then

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

Similarly,

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

\(\displaystyle \iint_D f(x,y)dA = \int_c^d dy \int_{g(y)}^{f(y)} f(x,y)dx \)

x is bounded by \(\displaystyle 0 \le x \le 1 \) and y is bounded by \(\displaystyle x^2 \le y \le x \)

Following the form from the above definitions we find that,

\(\displaystyle Area = \int_0^1 dx \int_{x^2}^{x} dy \)

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,

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

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

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

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

\(\displaystyle \int_0^{ 2 \pi } d \theta \int_0^4 dr = \int_0^4 dr \int_0^{ 2 \pi } d \theta = 8 \pi \)

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__

Consider the following problem1. 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 \(\displaystyle dA =dxdy \)

Let's follow the steps,Evaluate \(\displaystyle \iint_T (x-3y)dA \) where T is the triangle with the verices \(\displaystyle (0,0), (a,0), (0,b) \)

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

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,

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

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 \(\displaystyle y = mx + b \)

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

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

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

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

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

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

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

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

Let's look at the first part,

Note that: \(\displaystyle I = I_1 + I_2 \)

\(\displaystyle I_1 = \int_0^a dx \int_0^{ - \frac{b}{a} x + b } xdy \)

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!

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

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

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

Now for the second part of the integral we have,

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

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

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

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

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

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.

We do this when the integral, in its present form, is relatively hard to compute.When we are given a problem such that

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

and we change it to

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

it is known as a change of bounds.

Consider the following problem

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

If we set up the integral as given we get,

\(\displaystyle \int_0^1 xdx \int_{x^2}^x \frac{e^y}{y} dy \)

To put it simply, there is no good way to integrate \(\displaystyle \int_{x^2}^x \frac{e^y}{y} dy \). 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 \(\displaystyle y=f(x) \) into functions such that \(\displaystyle h(y) = x \) 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,

\(\displaystyle x^2 \le y \le x \) leads to \(\displaystyle y = x^2 \le y \le y= x \)

\(\displaystyle y =x^2 \to x = \sqrt{y} \) and \(\displaystyle y=x \to x=y \)

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 \(\displaystyle 0 \le y \le 1 \)

And on that interval, \(\displaystyle y \le \sqrt{y} \)

Thus,

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

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. \(\displaystyle y=x^2 \to x= \sqrt{y} \) ) but this is not the case. If we didn't have to idenitfy any further we would think,

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

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,

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

Let \(\displaystyle U = 1-y \) and \(\displaystyle dV = e^y dy \)

Such that \(\displaystyle du = -dy \) and \(\displaystyle V =e^y \)

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

Bellow are some homework questions for multiple integration.

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

\(\displaystyle \iint_S (sinx + cosy)dA \)

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

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

\(\displaystyle = \int_0^{ \frac{ \pi}{2}} [ \frac{ \pi}{2} sinx + 1] dx \)

\(\displaystyle = \pi \)

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

\(\displaystyle \iint_D lnx dA \)

Intersection of the line \(\displaystyle 2x+2y = 5 \) and the hyperbola \(\displaystyle xy=1 \) happens when

\(\displaystyle 2x^2 -5x + 2 = (2x-1)(x-2)=0\)

The intersections are at \(\displaystyle x = \frac{1}{2} \) and \(\displaystyle x = 2 \)

\(\displaystyle \iint_D lnx dA \)

\(\displaystyle \int_{ \frac{1}{2}}^2 lnxdx \int_{\frac{1}{x}}^{ \frac{5}{2} - x } dy \)

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

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

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

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

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

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

\(\displaystyle = \frac{33}{8}ln2 - \frac{45}{16}\)

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

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