- #1
Fractal20
- 74
- 1
Homework Statement
Find the volume of the region D in R^3 which is inside the sphere x^2 + y^2 + z^2 = 4 and also inside the cone z = sqrt (x^2 + y^2)
Homework Equations
The Attempt at a Solution
So I decided that the best approach might be finding the area under the sphere and then subtracting the area under the cone. The first step then seemed to be to find the corresponding projection onto the x-y plane that both of these have. Since the cone defines z as the sqrt (x^2 + y^2) plugging this into the equation for the sphere yields x^2 + y^2 = 2. I took this to mean that the projection is the circle of radius two centered on the origin.
Now for the area under the sphere. In polar coords, r varies from 0 to 2, and ∅ from 0 to 2∏. Translating the equation for the sphere into polar coords I got z = sqrt (4 - r^2). So then altogether I have [itex]\int[/itex][itex]^{2\pi}_{0}[/itex][itex]\int[/itex][itex]^{2}_{0}[/itex]sqrt(4 - r^2)r dr d∅.
For the cone it seemed like cylindrical coords would be the way to go. I thought z varies from 0 to r, r from 0 to 2, and ∅ from 0 to 2∏. So I had the integrand [itex]\int[/itex][itex]^{2\pi}_{0}[/itex][itex]\int[/itex][itex]^{2}_{0}[/itex] [itex]\int[/itex][itex]^{r}_{0}[/itex] r dz dr d∅.
Then I would just subtract the two. I'm really rusty on this stuff so I'm unsure in particular about my limits in the integration and the projection space. Does this seem correct? Is there an easier approach? This is a question from an old placement exam. Thanks!