# Thread: Volume of a soild using double integralsand polar coordinates

1. ## Volume of a soild using double integralsand polar coordinates

Hello

I am trying to find the volume the region above the cone $z=\sqrt{x^2+y^2}$ and below the sphere $x^2+y^2+z^2=1$

I was told to consider the difference $z=\sqrt{x^2+y^2} - \sqrt{1-(x^2+y^2)}$

and to set the two equations equal to each other to get bounds

$\sqrt{x^2+y^2}=\sqrt{1-(x^2+y^2)}$

simplifies to

$x^2+y^2 = \frac{1}{2}$

so
$0 \leq \theta \leq 2\pi$

and
$0 \leq r \leq \frac{1}{\sqrt{2}}$

I can easily set up the integral from here. But I don't understand the geometry behind these steps. Could someone explain to me please?

2. ## Re: Volume of a soild using double integralsand polar coordinates

Reverse the difference...

$\sqrt{1 - (x^2 + y^2)} - \sqrt{x^2 + y^2}$

$\text{i.e. }\sqrt{1 - r^2} - r$

and see it as the range of the inner-most integral of three...

$\displaystyle{\int_0^{2 \theta} \int_0^{\frac{1}{\sqrt{2}}} \int_r^{\sqrt{1 - r^2}}\ r\ dz\ dr\ d\theta$

... i.e. at every step along every 1/root2 radius about the origin in the z=0 plane, you are measuring the distance from the point on the cone directly above you (where z = r) up to as far as the point on the sphere directly above that (where z = root(1 - r^2)).