## anonymous 5 years ago Use polar coordinates to set up the double integral that defines the volume of a solid above the cone z = sqrt(x^2+y^2) and below the sphere x^2+y^2+z^2 =1

1. anonymous

You're looking to form the double integral such that, over the domain of integration, the net volume will be (volume of sphere) - (volume of cone). You need to find the domain of integration first, which means finding the boundary of the set of all (x,y) pairs you'll be using. This is found by the arc of intersection between the two surfaces, since at this arc, they will meet (one is opening upward (cone), the other opening downward (sphere)). So you solve$\sqrt{x^2+y^2}=\sqrt{1-x^2-y^2}$since we want all (x,y) that satisfies the height where they intersect.

2. anonymous

If you do this, you end up with$2x^2+2y^2=1 \rightarrow x^2+y^2=\frac{1}{2}$which is a circle in the x-y plane of radius $\frac{1}{\sqrt{2}}$

3. anonymous

Is this making sense?

4. anonymous

Since you're looking for the volume over the entire region with polar coordinates, your limits will be:$0 \le \theta \le 2\pi$and$0 \le r \le \frac{1}{\sqrt{2}}$

5. anonymous

$V=\int\limits_{0}^{1/\sqrt{2}}\int\limits_{0}^{2 \pi}\sqrt{1-r^2}-r .rdr d \theta$

6. anonymous

Where the part under the square root is found by applying the definition of the transformation x=rcos theta and y = r sin theta to the surface of the sphere. The r is found similarly, but by transforming the equation for the cone. The area of integration, dxdy is transformed as r dr d theta. You just have a normal double integral now. Note that the integrand is independent of theta, so you can integrate that out immediately. $2 \pi \int\limits_{0}^{1/\sqrt{2}}r \sqrt{1-r^2} - r^2 dr$

7. anonymous

You can integrate this easily, once you make a sub. for the radicand: u=1-r^2 then du=-rdr --> rdr = -du and you have r sqrt(1-r^2) dr = -u^(1/2)du. I think you can take it from there. Remember to undo your substitution for u at the end so you can use the appropriate limits for r. Good luck :)