# Confirmation of Thoughts/Results on a Few Double Integration Problems

• Aug 5th 2012, 01:27 PM
Phugoid
Confirmation of Thoughts/Results on a Few Double Integration Problems
Hello there,

I'm tutoring somebody at the moment and I'm having some trouble with a few double integral questions that have cropped up in their past papers. I'm not sure whether or not there are errors in the questions that make them impossible to tackle or whether I'm just over thinking things, so I'm looking for some confirmation. I'll give you the question, my final answer and explain my issue, and hopefully you guys can clear things up.

1) Sketch the domain of integration and evaluate the integral

$\displaystyle I = \int_1^3 \int_{2y}^{0} (yx^2+xy^2) \, dx \, dy$

For this I can evaluate the integral and get an answer of $\displaystyle I = -\frac{3888}{15}$, but I can't find a way to sketch the region.

I've drawn a graph with the lines$\displaystyle y = 1 \,\,\, y = 3 \,\,\, x = 0 \,\,\, x = 2y$, but there is no region enclosed by these curves, so how can the region be sketched/specified? Also, if no such region exists, how is it possible to evaluate the integral?

2) Evaluate $\displaystyle \displaystyle I = \int_{1}^{2x} \int_0^2 \left(yx^3 + xy^2\right) \, dx \, dy$

Here I am able to evaluate the integral and arrive at the result:

$\displaystyle I = \frac{16}{3}x^3 + 8x^2 - \frac{8}{3}$.

However, I'm slightly concerned that I'm getting a function of $\displaystyle x$ as a result rather than a numerical result... is this correct?

Either I'm correct and I'm over-thinking it, or it is an erroneous question (any double integrals I've seen in the past have the variable limits on the inner integral and the numerical limits on the outer integral, hence giving a numerical result), OR there is another way to deal with double integrals with variable limits on the outer integral that I'm not yet aware of. Which is it?

3) The height ($\displaystyle z$) of the ceiling of an art gallery is defined by the surface:

$\displaystyle z = 5+xy$

The gallery wall is circular (in plan) and is defined by the equation:

$\displaystyle x^2 + y^2 = 36$

For $\displaystyle y > 0$, derive an integral expression for the area of the wall between the limits $\displaystyle 0 \leq x \leq 5$.

For this one I'm not sure where to get started at all. It might not even be a double integral, but some sort of line integral or something? Anybody have any hints?
• Aug 5th 2012, 02:21 PM
GJA
Re: Confirmation of Thoughts/Results on a Few Double Integration Problems
Hi, Phugoid. I'm a little confused about the region for problem #1 as well.

For problem #3 do they want the surface area of the ceiling along with the outer walls?

To find the surface area of the ceiling you can use the formula

$\displaystyle \int\int\sqrt{1+f_{x}^{2}+f_{y}^{2}}dA$,

where $\displaystyle f(x,y)=5+xy$. See Area - Wikipedia, the free encyclopedia about three quarters of the way down for this formula.
• Aug 5th 2012, 02:33 PM
Phugoid
Re: Confirmation of Thoughts/Results on a Few Double Integration Problems
Quote:

Originally Posted by GJA
Hi, Phugoid. I'm a little confused about the region for problem #1 as well.

Doesn't make any sense at all, does it?

Quote:

For problem #3 do they want the surface area of the ceiling along with the outer walls?

To find the surface area of the ceiling you can use the formula

$\displaystyle \int\int\sqrt{1+f_{x}^{2}+f_{y}^{2}}dA$,

where $\displaystyle f(x,y)=5+xy$. See Area - Wikipedia, the free encyclopedia about three quarters of the way down for this formula.
It is my interpretation of #3 that they want the area of the wall alone, given that the height of the wall is dependent on the value of $\displaystyle z$ along the curve $\displaystyle y = \sqrt{36-x^2}$ for $\displaystyle 0 \leq x \leq 5$.

Basically an "area under a curve" problem except that the area of projected onto the inside of a circle rather than a flat surface. If that is the case, I would probably proceed as follows:

$\displaystyle I = \int_S z \, ds\\ \newline=\int_S (5+xy) \, ds\\ \newline =\int_S (5+xy) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\\ \newline=\int_{x_1}^{x_2} (5+xy)\sqrt{1 + \left(\frac{x^2}{36 - x^2}\right)} \, dx\\ \newline =\int_{0}^{5} (5+xy)\sqrt{1 + \left(\frac{x^2}{36 - x^2}\right)} \, dx \\ \newline =\int_{0}^{5} (5+x\sqrt{36 - x^2})\sqrt{1 + \left(\frac{x^2}{36 - x^2}\right)} \, dx$

Anybody concur?

What are your thoughts on #2?
• Aug 5th 2012, 02:58 PM
HallsofIvy
Re: Confirmation of Thoughts/Results on a Few Double Integration Problems
Quote:

Originally Posted by Phugoid
Hello there,

I'm tutoring somebody at the moment and I'm having some trouble with a few double integral questions that have cropped up in their past papers. I'm not sure whether or not there are errors in the questions that make them impossible to tackle or whether I'm just over thinking things, so I'm looking for some confirmation. I'll give you the question, my final answer and explain my issue, and hopefully you guys can clear things up.

1) Sketch the domain of integration and evaluate the integral

$\displaystyle I = \int_1^3 \int_{2y}^{0} (yx^2+xy^2) \, dx \, dy$

For this I can evaluate the integral and get an answer of $\displaystyle I = -\frac{3888}{15}$, but I can't find a way to sketch the region.

I've drawn a graph with the lines$\displaystyle y = 1 \,\,\, y = 3 \,\,\, x = 0 \,\,\, x = 2y$, but there is no region enclosed by these curves, so how can the region be sketched/specified? Also, if no such region exists, how is it possible to evaluate the integral?

I don't know why you say that. y= 1 and y= 3 are horizontal lines. x= 0 crosses those lines at (0,1) and (0,3), of course. And x= 2y crosses y= 1 at x= 2 and y= 3 at x= 6. That is, the region is the trapezoid with vertices at (0,1), (2,1), (6, 3), and (0,3). Looks straightforward to me.

Quote:

2) Evaluate $\displaystyle \displaystyle I = \int_{1}^{2x} \int_0^2 \left(yx^3 + xy^2\right) \, dx \, dy$

Here I am able to evaluate the integral and arrive at the result:

$\displaystyle I = \frac{16}{3}x^3 + 8x^2 - \frac{8}{3}$.

However, I'm slightly concerned that I'm getting a function of $\displaystyle x$ as a result rather than a numerical result... is this correct?

Either I'm correct and I'm over-thinking it, or it is an erroneous question (any double integrals I've seen in the past have the variable limits on the inner integral and the numerical limits on the outer integral, hence giving a numerical result), OR there is another way to deal with double integrals with variable limits on the outer integral that I'm not yet aware of. Which is it?
Technically, it is not an error. The first "x", as a variable of integration, is a "dummy" variable and there is no "x" in the problem after you have done the first integral so you can legally use "x" as a limit of integration in the second integral. Of course, it would be easier to understand if we used a different dummy variable- say $\displaystyle \int_0^{2x}\int_0^2 yt^3+ ty^2 dt dy$.

Quote:

3) The height ($\displaystyle z$) of the ceiling of an art gallery is defined by the surface:

$\displaystyle z = 5+xy$

The gallery wall is circular (in plan) and is defined by the equation:

$\displaystyle x^2 + y^2 = 36$

For $\displaystyle y > 0$, derive an integral expression for the area of the wall between the limits $\displaystyle 0 \leq x \leq 5$.

For this one I'm not sure where to get started at all. It might not even be a double integral, but some sort of line integral or something? Anybody have any hints?
No, its a pretty standard integral over an area. The only result of the last limits is that instead of integrating, with respect to x, from -6 to 6 (covering the entire circle), you integrate from 0 to 5. For each x, of course, $\displaystyle y= \pm\sqrt{36- x^2}$. The integral is
$\displaystyle \int_{x=0}^5\int_{y=-\sqrt{36- x^2}}^{\sqrt{36- x^2}} 5+ xy dydx$
• Aug 5th 2012, 03:05 PM
GJA
Re: Confirmation of Thoughts/Results on a Few Double Integration Problems
If it's just the wall under the circular arc, you can do a line integral by parameterizing the arc of the circle you're integrating over.

I'm a little confused on #2 as well. Maybe I'm forgetting something, or maybe they do want your answer in terms of x, but I can't say I recall an example I've worked like that recently.

Update: For some reason the site is not letting me post new things, so I am forced to respond to the threads by updating my previous post.

First, I agree with your most recent post, Phugoid. It seems to me that to get the inequalities on x are not correct.

Second, I don't think you want to do the area of the wall problem as a double integral; that would give you the volume of the room. I think doing a line integral by parameterizing the arc you're interested in is the way to go.
• Aug 5th 2012, 03:26 PM
Phugoid
Re: Confirmation of Thoughts/Results on a Few Double Integration Problems
Quote:

Originally Posted by HallsofIvy
I don't know why you say that. y= 1 and y= 3 are horizontal lines. x= 0 crosses those lines at (0,1) and (0,3), of course. And x= 2y crosses y= 1 at x= 2 and y= 3 at x= 6. That is, the region is the trapezoid with vertices at (0,1), (2,1), (6, 3), and (0,3). Looks straightforward to me.

I agree that there is a region bounded by those lines, but because $\displaystyle 2y \leq x \leq 0$ I thought it should be the case that the lower bound of the region should by the line $\displaystyle x = 2y$ rather than the line $\displaystyle x = 0$ and vice versa.

For me, the region you describe is $\displaystyle 1 \leq y \leq 3 \,\,\,\, 0 \leq x \leq 2y$ and not $\displaystyle 1 \leq y \leq 3 \,\,\,\, 2y \leq x \leq 0$

Quote:

Technically, it is not an error. The first "x", as a variable of integration, is a "dummy" variable and there is no "x" in the problem after you have done the first integral so you can legally use "x" as a limit of integration in the second integral. Of course, it would be easier to understand if we used a different dummy variable- say $\displaystyle \int_0^{2x}\int_0^2 yt^3+ ty^2 dt dy$.
I understand, but I was just concerned by the fact that the result of the whole integral is a function of $\displaystyle x$ rather than a numerical value. I've never seen a double integral like that before - usually the "variable" limits are on the inner integral, and the numeral limits ar eon the outer integral, hence always resulting in a numerical result in the evaluation of the integral.

Quote:

No, its a pretty standard integral over an area. The only result of the last limits is that instead of integrating, with respect to x, from -6 to 6 (covering the entire circle), you integrate from 0 to 5. For each x, of course, $\displaystyle y= \pm\sqrt{36- x^2}$. The integral is
$\displaystyle \int_{x=0}^5\int_{y=-\sqrt{36- x^2}}^{\sqrt{36- x^2}} 5+ xy dydx$
Not sure where you got that double integral from. It states that you're supposed to evaluate it for $\displaystyle y > 0$ so how does the lower limit of your inner integral come into place?

Also do you agree with my working for that question in post #3?
• Aug 6th 2012, 01:59 AM
Phugoid
Re: Confirmation of Thoughts/Results on a Few Double Integration Problems
Quote:

Originally Posted by HallsofIvy
.

BUMP!

Anyone?
• Aug 6th 2012, 02:45 AM
Prove It
Re: Confirmation of Thoughts/Results on a Few Double Integration Problems
For the first one, the property \displaystyle \displaystyle \begin{align*} \int_b^a{f(x)\,dx} = -\int_a^b{f(x)\,dx} \end{align*} should help you...
• Aug 6th 2012, 04:06 AM
Phugoid
Re: Confirmation of Thoughts/Results on a Few Double Integration Problems
Quote:

Originally Posted by Prove It
For the first one, the property \displaystyle \displaystyle \begin{align*} \int_b^a{f(x)\,dx} = -\int_a^b{f(x)\,dx} \end{align*} should help you...

I agree. I'm satisfied with that one now.

For #2 I'd just like for somebody to confirm that it's OKAY to get a function of x for an answer to the integral - or, if not, then how do you deal with that integral to GET a numerical answer. And for #3 could anybody confirm that they agree with my working?

I tried a simpler test case for #3 - i.e. if the ceiling was just a constant 5, say, and we were interested in the area of the wall for 0 <= x <= 6, then my integral comes out to be $\displaystyle 15 \pi$, and it is easy to verify that this would be the area of a wall that's composed of a quarter circle of radius 6 and height 5. So on that basis, I'm fairly confident of my result - if anybody could corroborate, I would be happy!
• Aug 6th 2012, 04:09 AM
Prove It
Re: Confirmation of Thoughts/Results on a Few Double Integration Problems
Quote:

Originally Posted by Phugoid
I agree. I'm satisfied with that one now.

For #2 I'd just like for somebody to confirm that it's OKAY to get a function of x for an answer to the integral - or, if not, then how do you deal with that integral to GET a numerical answer. And for #3 could anybody confirm that they agree with my working?

I tried a simpler test case for #3 - i.e. if the ceiling was just a constant 5, say, and we were interested in the area of the wall for 0 <= x <= 6, then my integral comes out to be $\displaystyle 15 \pi$, and it is easy to verify that this would be the area of a wall that's composed of a quarter circle of radius 6 and height 5. So on that basis, I'm fairly confident of my result - if anybody could corroborate, I would be happy!

I expect with 2, you need to switch the order of integration (note you won't need to change your bounds, check the region with a sketch if you must).
• Aug 6th 2012, 05:04 AM
Phugoid
Re: Confirmation of Thoughts/Results on a Few Double Integration Problems
Quote:

Originally Posted by Prove It
I expect with 2, you need to switch the order of integration (note you won't need to change your bounds, check the region with a sketch if you must).

So you're saying I can go:

$\displaystyle \displaystyle I = \int_{1}^{2x} \int_0^2 \left(yx^3 + xy^2\right) \, dx \, dy = \int_0^2 \int_{1}^{2x} \left(yx^3 + xy^2\right) \, dy \, dx$

Can you explain why I don't need to change the bounds in this case? I thought it was only for the integration of a Rectangle that it would not be necessary to change the bounds in a change of order of integration?
• Aug 6th 2012, 05:18 AM
Prove It
Re: Confirmation of Thoughts/Results on a Few Double Integration Problems
Quote:

Originally Posted by Phugoid
So you're saying I can go:

$\displaystyle \displaystyle I = \int_{1}^{2x} \int_0^2 \left(yx^3 + xy^2\right) \, dx \, dy = \int_0^2 \int_{1}^{2x} \left(yx^3 + xy^2\right) \, dy \, dx$

Can you explain why I don't need to change the bounds in this case? I thought it was only for the integration of a Rectangle that it would not be necessary to change the bounds in a change of order of integration?

I mean because it appears that your bounds are ALREADY written as though you're summing up horizontal strips. If you had to change from horizontal to vertical, then you would have to change the bounds...
• Aug 6th 2012, 05:30 AM
Phugoid
Re: Confirmation of Thoughts/Results on a Few Double Integration Problems
Quote:

Originally Posted by Prove It
I mean because it appears that your bounds are ALREADY written as though you're summing up horizontal strips. If you had to change from horizontal to vertical, then you would have to change the bounds...

Ah, I see what you mean. The integration limits are written as if they're summing up the horizontal strips (i.e. $\displaystyle \,dy \,dx$), but for some reason the question has it written $\displaystyle \, dx \,dy$ instead. So it's just a case of correct that by writing:

$\displaystyle \displaystyle \int_1^{2x} \int_0^2 (yx^3+xy^2) \, dy \, dx$

I'm not really changing the order of integration, I'm just correcting the initial expression because it was written inappropriately.