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Calculus/Calculating the volume of R^3 lay between the sphere and cilinder

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Question
Dear Abe,

So you have a 3d space where a sphere is present with the equation

x+y+z=4a

z=sqrt(4a-x-z)


A cilinder cuts a certain volume out of this sphere, it's up to me to calculate this volume. The cilinder has cartesian coordinates

x+y=2ay

I can recognize the equation of a circle (x-x0)+(y-y0)=r
I rewrite it in this form to find the center of the cilinder

x+y-2ay+a=a

(x-x0)+(y-a)=a

so center is (0,a) and radius is a

I go onto polar coordinate system:

p=2apsin&

p(max)=2asin&
p(min)=0

&(max)=pi
&(min)=0

So I rewrite the integral

2*integral(0 to pi)integral(0 to 2asin&)sqrt(4a-p)pdpd&
-1*integral(0 to 2pi)integral(0 to 2asin&)sqrt(4a-p)d(4a-p)d&

This gives me

-1*integral(0 to 2pi)[(4a-p)^(3/2)/(3/2)]from p=0 to p=2asin&

Up to this I know the calculations are correct. However the following things aren't very clear to me:

Why is that z=sqrt(4a-x-y) only half of the volume is?
Why are the borders of the integral to & readjusted to 2pi during integration?

So now I need to integrate further to &, but there is where I get lost

The result I need to get is (16a^3/6)*(pi-(4/3))

Thanks,
Bob

Answer
Hello Bob,

z=sqrt(4a-x-y) is only half of the volume since the cylinder and the sphere
also intersect on the other side (z<0) of the xy-plane.

Your first integral is correct: 2*integral(0 to pi)integral(0 to 2asin&)sqrt(4a-p)pdpd&
which gives: (16/9)*a^3*(3*Pi-4)...I think the "6" in your answer should be a "3"...

Abe

Calculus

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Abe Mantell

Expertise

Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!

Experience

Over 15 years teaching at the college level.

Organizations
NCTM, NYSMATYC, AMATYC, MAA, NYSUT, AFT.

Education/Credentials
B.S. in Mathematics from Rensselaer Polytechnic Institute
M.S. (and A.B.D.) in Applied Mathematics from SUNY @ Stony Brook

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