Volume of solid using cylindrical cordinates

[aron101782]

Find the volume of the solid that the cylinder r=acost cuts out of the sphere of radius a.

I am getting two different solutions. When integrating

4I(0 pi/2)I(0 acost)I(0 (a^2-r^2)^(1/2)) r dz dr dt = 1.205
(correct in the solutions manual)

I(-pi/2 pi/2)I(0 acost)I(-(a^2-r^2)^1/2 (a^2-r^2)1/2) r dz dr dt = 2.09
(incorrect in the solutions manual)

if if this is true. I would like to know why it is incorrect to integrate through the entire region. And correct to integrate through the sub region and multiply by 4.

 

[tkhunny]

Find the volume of the solid that the cylinder r=acost cuts out of the sphere of radius a.

I am getting two different solutions. When integrating

4I(0 pi/2)I(0 acost)I(0 (a^2-r^2)^(1/2)) r dz dr dt = 1.205
(correct in the solutions manual)

I(-pi/2 pi/2)I(0 acost)I(-(a^2-r^2)^1/2 (a^2-r^2)1/2) r dz dr dt = 2.09
(incorrect in the solutions manual)

if if this is true. I would like to know why it is incorrect to integrate through the entire region. And correct to integrate through the sub region and multiply by 4.

1) Where is the sphere centered?
2) Is the result independent of a?
3) If [0,pi/2] is 1/2 of the region, then isn't [-pi/2,pi/2] only twice the size? Are you sure 4 ==> 1 is the correct change of external factor?

 

[aron101782]

1) Where is the sphere centered?
2) Is the result independent of a?
3) If [0,pi/2] is 1/2 of the region, then isn't [-pi/2,pi/2] only twice the size? Are you sure 4 ==> 1 is the correct change of external factor?

It's a simple question yeah I suppose I will have to answer for myself thanks to people like you.

 

[Dr.Peterson]

It's a simple question yeah I suppose I will have to answer for myself thanks to people like you.

Well, we need the answers to the questions you were asked in order to help you.

You didn't state the problem clearly, so that we can be sure what region you are trying to integrate over. Your 4 suggests that my interpretation is wrong.

The problem involves a radius a, but your answers (even the one you say the book provided) are independent of a; so something is going on there. It would be very helpful to see some of your work, so we can get an idea where you are going wrong. Maybe you read the wrong answer in the book!

I, for one, had been waiting to see the answers, so I can help if needed. Apparently I won't get to do that, because you choose not to help others help you.

 

[aron101782]

Well, we need the answers to the questions you were asked in order to help you.

You didn't state the problem clearly, so that we can be sure what region you are trying to integrate over. Your 4 suggests that my interpretation is wrong.

The problem involves a radius a, but your answers (even the one you say the book provided) are independent of a; so something is going on there. It would be very helpful to see some of your work, so we can get an idea where you are going wrong. Maybe you read the wrong answer in the book!

I, for one, had been waiting to see the answers, so I can help if needed. Apparently I won't get to do that, because you choose not to help others help you.

Ya whatever you say doctor. But your system has an error. It won't let me upload a photo of my solution. Here it is typed without a skeetch.

So we have a sphere of radius a centered at the origin. And a cylinder r=acost.

For the equation of the sphere
x^2+y^2+z^2=a^2
z=+-(a^2-r^2)^1/2

And plotting points for r=acost see that this is a cylinder centered at 0,a/2,0

We want to find the volume of the cylinder bound by the sphere

I(-pi/2, pi/2)I(0, acost)I(-(a^2-r^2)^1/2, (a^2-r^)^1/2)r dzdrdt
I(-pi/2, pi/2)I(0, acost)2r(a^2-r^2)^1/2 drdt
2/3 I(-pi/2, pi/2) a^3-a^3sin^3^t dt
2a^3/3( pi-I(-pi/2, pi/2)sint(1-cos^2t)dt)
2a^3/3(pi+I(0,0)1-u^2 du)
At the previous line there appears to be a problem
2a^3 pi/3
INCORRECT

However when setting the bounds of integration to be in the first quadrant we have
4 I(0, pi/2)I(0, acost)I(0, (a^2-r^2)^1/2) r dzdrdt
4 I(0, pi/2)I(0, acost) r(a^2-r^2)^1/2 drdt
4a^3/3 I(0, pi/2)1-sin^3t dt
4a^3/3(pi/2-I(0, pi/2)sint(1-cos^2t)dt
4a^3/3(pi/2-I(0,1)1-u^2 du)
4a^3/3(pi/2-(1-1/3))
4a^3/3(pi/2-2/3)
2a^3(3pi-4)/9
CORRECT

 

[Otis]

… [the forum] system has an error. It won't let me upload a photo of my solution …

Hi. The vBulletin forum software setup here contains a number of bugs and restrictions. We all deal with it, ha.

It's better to host large images at imgbb, instead. See this notice, for more information. :cool:

 

[Dr.Peterson]

I(-pi/2, pi/2)I(0, acost)I(-(a^2-r^2)^1/2, (a^2-r^)^1/2)r dzdrdt
I(-pi/2, pi/2)I(0, acost)2r(a^2-r^2)^1/2 drdt
2/3 I(-pi/2, pi/2) a^3-a^3sin^3^t dt
2a^3/3( pi-I(-pi/2, pi/2)sint(1-cos^2t)dt)
2a^3/3(pi+I(0,0)1-u^2 du)
At the previous line there appears to be a problem
2a^3 pi/3
INCORRECT

However when setting the bounds of integration to be in the first quadrant we have
4 I(0, pi/2)I(0, acost)I(0, (a^2-r^2)^1/2) r dzdrdt
4 I(0, pi/2)I(0, acost) r(a^2-r^2)^1/2 drdt
4a^3/3 I(0, pi/2)1-sin^3t dt
4a^3/3(pi/2-I(0, pi/2)sint(1-cos^2t)dt
4a^3/3(pi/2-I(0,1)1-u^2 du)
4a^3/3(pi/2-(1-1/3))
4a^3/3(pi/2-2/3)
2a^3(3pi-4)/9
CORRECT

Thanks for responding. Now I can see the problem.

In the first version, you are letting r vary from -pi/2 to pi/2, so that sin(t) varies from -1 to 1. But when you go from I(a^2-r^2)^1/2 dr to 2/3 a^3-a^3sin^3(t), you assumed that (1 - cos^2(t))^(3/2) = (sin^2(t))^(3/2) = sin^3(t). That is true only for positive values! Really, it should be |sin^3(t)|.

To put it another way, your substitution, u = cos(t), was not one-to-one over the required interval. When you change variables in the limits of integration, you need to do it in two pieces (-pi/2 to 0, and 0 to pi/2) to get it right.

Doing only one half and doubling the result avoids this difficulty, which is why it is a good idea!

 

[aron101782]

Yes that seems to make sense. But it is still unclear to me. How exactly do I set up the integral in 2 parts to arrive at the correct solution.

-Aron

 

[Dr.Peterson]

Yes that seems to make sense. But it is still unclear to me. How exactly do I set up the integral in 2 parts to arrive at the correct solution.

-Aron

As I said, make the sum of integrals from -pi/2 to 0, and 0 to pi/2; and in each one, watch the signs in your work. Give it a try.