0
$\begingroup$

Find the volume of the region: $$\iiint(x^2+y^2+z^2)dV $$ where $R$ is the region above the cone $z = a\sqrt{x^2+y^2}$ and inside the sphere $x^2+y^2+z^2=b^2$.

I am trying to use spherical coordinates to solve this with the bounds of rho as 0 to b and the bounds of theta from 0 to 2 Pi but I am not sure what the bounds of phi would be when the cone and the sphere intersect.

I also know the integral expression should be $\rho^4\sin\phi d\rho d\phi d\theta$

$\endgroup$
0
$\begingroup$

Hint. Where the two surfaces intersect, we have that $(1+a^2)(x^2+y^2)=b^2,$ or $$\rho=\frac{b}{\sqrt{1+a^2}}.$$ This gives the limits for $\rho.$ So the radius doesn't go all the way to $b.$

Thus in cylindrical coordinates the integral simplifies to $$2π\int_0^{b/\sqrt{1+a^2}} \rho^3+\frac{a^3}{3}\rho^4-\frac13\rho(a^2-\rho^2)\sqrt{a^2-\rho^2}\,\mathrm d\rho,$$ which may be evaluated easily. The first part is a polynomial in $\rho.$ For the second part take $\rho=a\sin\psi.$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So would the answer be $(-2πb^5)* a/\sqrt{a^2+1}-1/5$ $\endgroup$ – FreshDough Apr 8 at 18:22
  • $\begingroup$ @FreshDough That's odd. You should get a nonnegative answer. Check for an error. $\endgroup$ – Allawonder Apr 8 at 23:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.