# Thread: Double Integral in Polar Coordinates

1. ## Double Integral in Polar Coordinates

Use polar coordinates to find the volume of the solid above the cone z = (x^2+y^2)^1/2 and below the hemisphere
z =( 8 − x^2 − y^2)^1/2.

2. Originally Posted by crwhd4
Use polar coordinates to find the volume of the solid above the cone z = (x^2+y^2)^1/2 and below the hemisphere
z =( 8 − x^2 − y^2)^1/2.
You need to find

$V = \int \, \int_{R_{xy}} \sqrt{8 - x^2 - y^2} - \sqrt{x^2 + y^2} \, dx \, dy$

where $R_{xy}$ is the region inside the circle $x^2 + y^2 = 4$.

Note: $x^2 + y^2 = 8 - x^2 - y^2 \Rightarrow x^2 + y^2 = 4$.

Make the switch to polar coordinates:

$V = \int_{\theta = 0}^{\theta = 2 \pi} \int_{r = 0}^{r = 2} (\sqrt{8 - r^2} - r) \, r \, dr \, d\theta$

$= \int_{r = 0}^{r = 2} (\sqrt{8 - r^2} - r) \, r\, dr \cdot \int_{\theta = 0}^{\theta = 2 \pi} d \theta$.

Now do the simple integrations.

3. This is probably a stupid question, but how is it that an iterated integral becomes a product of integrals? Or is that the case in this one only because the inner integral will yield a constant?

4. Originally Posted by rualin
This is probably a stupid question, but how is it that an iterated integral becomes a product of integrals? Or is that the case in this one only because the inner integral will yield a constant?
The variables are seperable. In general:

$\int \int f(x) \, g(y) \, dx \, dy = \int f(x) \, dx \cdot \int g(y) \, dy$