Use polar coordinates to find the volume of the given solid bounded by the paraboloid $z=1+2x^2+2y^2$ and the plane $z=7$ in the first octant.
I did it. Is that right ?
$$\int_0^{\pi \over 2} \int_0^{\sqrt{3}}(7-(1+2r^2))r dr d\theta = \frac{9\pi}{4} $$
Thanks
Best Answer
Answer:
You are almost right except that the integrand is ($7-(1+2r^2)$) = $(6-2r^2)$