Multi-variable calc question Use spherical coordinates to evaluate the triple integral?

In spherical coordinates the value x² + y² + z² = r². This gives the following spherical integral:

∫∫∫r² * r²sin(θ)drdθdφ

This is easier done by realizing that since you have spherical symmetry. Thus each shell of radius r² creates a volume of r² * A * dr, since the surface area of a sphere is 4πr², this gives:

∫r² * 4πr²dr, from r = 0 to r = 10 (i.e. r² = 100)

-->

4π∫r⁴dr = 4π * r⁵/5 + C, since r = 0 gives 0 we get:

4π * 10⁵/5 = 400,000π / 5 = 80,000π

If you carry out the first two integrals:

∫∫sin(θ)dθdφ, from θ = -π/2 to θ = +π/2 and φ = 0 to φ = 2π

you will get the same result because this integral equals 4π and thus you end up with:

4π∫r²r²dr, from r = 0 to r = 10

Ok so

(x,y,z) = (p, theta, thi) from rectangular to spherical

p^2=x^2+y^2+z^2

p = your radius

100=x^2+y^2+z^2

so

p^2=100

p=10

now in spherical coordinates your limits for theta will always be 0 to 2pi and thi will always be from

0 to pi (unless otherwise specified)

every time you integrate in spherical coordinates, you use p^2*sin(thi)*d(theta)d(thi)

/ = integral

so your volume will be

///p^2*sin(thi)*d(theta)d(thi), where

thi is from 0 to pi

theta is from 0 to 2pi

p is from 0 to 10