# Thread: Triple integral (spherical coordinate)

1. ## Triple integral (spherical coordinate)

find the volume of the solid D that lies above the cone z = (x^2 + y^2)^1/2
and below the sphere z = (x^2 + y^2 + z^2)

i've done the integration until i need to substitute cos phi = u..
however.. i dont know to change the range..

2. ## Re: Triple integral (spherical coordinate)

If $u = \cos\phi$ then $u = \cos0 = 1$ when $\phi=0$, and $u = \cos(\pi/4) = 1/\sqrt2$ when $\phi = \pi/4.$ That gives $V = -\frac{2\pi}3\int_1^{1/\sqrt2}u^3\,du.$ From there, the best thing to do is probably to switch the limits of integration, which changes the sign of the integral and leads to $V = \frac{2\pi}3\int^1_{1/\sqrt2}u^3\,du.$ That should be easy enough to integrate.