volume, double integral

[mathstresser]

Use polar coordinates to find the volume of the given solid.
Inside the sphere x^2 + y^2 + z^2= 16 and outside the cylinder x+2+y^2=4.

I changed the sphere so it is r^2 + z^2=16 and the cylinder to r^2=4.
z^2=16-r^2


∫(p1,p2) ∫(p3,p4) [(16-r^2)^(1/2) rdrdx

Should p3 and p4 be (-4,-2) and (2,4)?

should p1 and p3 be (0,2pi)?

If you couldn't tell, these really confuse me.

18

 

[galactus]

You could find the volume of the sphere and then the cylinder and subtract.

The radius of the sphere is 4. Find the volume in 1 octant and multiply by 8.

\(\displaystyle \L\\8\int_{0}^{\frac{\pi}{2}}\int_{0}^{4}r\sqrt{16-r^{2}}drd{\theta}\)


Should your cylinder be \(\displaystyle x^{2}+y^{2}=4\)?. You have \(\displaystyle x+2+y^{2}=4\)

 

[mathstresser]

Thanks!