Hi

the question is

Use polar coordinates to find the volume of the solid which is under the paraboloid $\displaystyle z=x^{2}+y^{2}$and above the disk $\displaystyle x^{2}+y^{2}\leq9$ in the plane $\displaystyle z=0$.

thanks(Wink)

- Jan 26th 2010, 07:11 AMrpatelQuestion to do with volume of a solid between a paraboloid and a plane
Hi

the question is

Use polar coordinates to find the volume of the solid which is under the paraboloid $\displaystyle z=x^{2}+y^{2}$and above the disk $\displaystyle x^{2}+y^{2}\leq9$ in the plane $\displaystyle z=0$.

thanks(Wink) - Jan 26th 2010, 09:15 AMshawsend
If $\displaystyle z=x^2+y^2$ then $\displaystyle z=f(r,t)=r^2$ so we integrate over the circle $\displaystyle x^2+y^2=9$ as the radius goes from 0 to 3 and all the way round the circle or just 4 times the part in the first quadrant since f is symmetrical:

$\displaystyle V=4\int_0^{\pi/2}\int_0^3 r f(r,t) dr dt$

and you can check that by just calculating the volume inside the paraboloid via shells and then subtract that from just the cylinder it's housed in:

$\displaystyle \pi (3)^2 9-\pi\int_0^9 ydy=4\int_0^{\pi/2}\int_0^3 r f(r,t) dr dt$

half-in and half-out right?