Calculus Revision Question: Double Integral

Forum for the GRE subject test in mathematics.
red_apricot
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Joined: Mon Sep 29, 2014 2:04 am

Calculus Revision Question: Double Integral

Postby red_apricot » Mon Sep 29, 2014 2:10 am

Hi,

I've been revising multivariate calculus for GRE and came across the following exercise:
Find the volume of a body, whose (x,y,z) coordinates satisfy
x^2+y^2+z^2 <= 2cz, x^2+y^2 >= 2az, 0 < a < c <= 2a.

The answer the book provides is (4pi/3)(c-a)^3. Is this indeed a correct answer?
I can't find any bugs in my solution, and my answer does not match this.

Thanks.

DDswife
Posts: 89
Joined: Thu Aug 14, 2014 5:29 pm

Re: Calculus Revision Question: Double Integral

Postby DDswife » Tue Sep 30, 2014 8:32 pm

Did you calculate the volume that is inside the sphere and outside the paraboloid

red_apricot
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Joined: Mon Sep 29, 2014 2:04 am

Re: Calculus Revision Question: Double Integral

Postby red_apricot » Tue Sep 30, 2014 10:03 pm

Ah thanks, now it works. It appears when I was switching to polar coordinates I mistakenly took \sqrt{2a(c-a)} as the upper limit for {\bf{r}} instead of 2\sqrt{a(c-a)}.
As a bonus for those who want to warm-up: the book contains also the case c >= 2a, for which the answer is also the same as above.

DDswife
Posts: 89
Joined: Thu Aug 14, 2014 5:29 pm

Re: Calculus Revision Question: Double Integral

Postby DDswife » Tue Sep 30, 2014 11:22 pm

I couldn't figure out what this condition was for. But thanks.

red_apricot
Posts: 4
Joined: Mon Sep 29, 2014 2:04 am

Re: Calculus Revision Question: Double Integral

Postby red_apricot » Thu Oct 02, 2014 12:01 am

Now I have another question concerning surface area:
Find the area of that part of the sphere x^2+y^2+z^2 = a^2 for which a(a+x) <= y^2.
The answer I've got is 2a^2\times(pi-2), but the book has 4a^2. I parametrize the surface as {-\sqrt{a^2-r^2}, rcos(phi), rsin(phi)} for \phi \in [0,pi/4].
Which one is correct?

DDswife
Posts: 89
Joined: Thu Aug 14, 2014 5:29 pm

Re: Calculus Revision Question: Double Integral

Postby DDswife » Thu Oct 02, 2014 10:30 am

I haven't reviewed Multivariable Calculus yet. But, if you post what you did, or send it to me, I should be able to at least check your work for mistakes. Or maybe someone else will.
Last edited by DDswife on Wed Nov 01, 2017 6:40 pm, edited 3 times in total.

red_apricot
Posts: 4
Joined: Mon Sep 29, 2014 2:04 am

Re: Calculus Revision Question: Double Integral

Postby red_apricot » Wed Oct 08, 2014 1:59 am

Now that we've got \TeX here, here is one exercise on volumes:
Find the volume of a body, bounded by the following surfaces
\[<br />(x/a+y/b+z/c)^3 = \sin\Bigl(\frac{\pi(x/a+y/b)}{x/a+y/b+z/c}\Bigr), x = 0, y = 0, z = 0<br />\]
where x,y,z > 0 and all the parameters a,b,c positive.
I switched to polar coordinates
\[<br />x = ar\cos^2\phi\cos^2\psi, y = br\cos^2\phi\sin^2\psi, z = cr\sin^2\phi<br />\]
with Jacobian being 4abcr^2\cos^3\phi\sin\phi\cos\psi\sin\psi
and my answer was \frac{abc}{3\pi}
But the book lists \frac{1}{3}\pi abc(\frac{1}{\pi}-\frac{1}{\pi^3}).
Here because of x,y,z > 0 the limits of integration for both \phi and \psi were [0,\pi/2].
My question is, what am I missing in my solution?
Last edited by red_apricot on Thu Oct 09, 2014 12:54 am, edited 1 time in total.

DDswife
Posts: 89
Joined: Thu Aug 14, 2014 5:29 pm

Re: Calculus Revision Question: Double Integral

Postby DDswife » Wed Oct 08, 2014 3:17 pm

Yesterday I posted something about your first question, but it's not showing here.

What I said is that we can calculate that in the x=0 plane by using the washers method. This is, to my taste, the easiest way to solve it.

I said too that I found out why the condition c<= 2a. I used a = 1, and c =2, and then c =3. I noticed that, in the first case, the center of the sphere (circle in x=0) and the intersection point (between the circle and the parabola) are at the same height (z). Otherwise, the intersection point is in the top quarter of the circle.




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