To Find: use cylindrical coordinates to find the volume of a solid that is enclosed between the surface $r^2+z^2=20$ and the surface $z=r^2$:
$$\iiint_R f(x,y,z)\,dv=\iiint_R f(r\cos\theta,r\sin\theta,z)r\,dV$$
.R is part of the cylinder $$x^2+y^2 \le R^2$$
bounded by $$x^2+y^2+z^2=20$$
$$z=x^2+y^2$$
$$z+z^2=20, z=4=x^2+y^2$$
take part of the cylinder $$\iiint_G\,dV$$
$$G=\left \{ (x,y,z):0 \le x^2+y^2 \le 4, x^2+y^2 \le z \le (20-x^2-y^2)^\frac{1}{2} \right \}$$
$$G=\left \{ (r,\theta,z): 0 \le r \le 2,0 \le \theta \le 2\pi, r^2 \le z \le (20-r^2)^\frac{1}{2} \right \}$$
$$Volume=\iiint_G \,dV=\int_0^2\int_0^{2\pi}\int_{r^2}^{({20-r^2})^{\frac{1}{2}}} r\,dz\,d\theta\,dr$$
$\endgroup$ 1But where to from here?
1 Answer
$\begingroup$You are almost there. Just keep going.
$$Volume=\iiint_G \,dV=\int_0^2\int_0^{2\pi}\int_{r^2}^{({20-r^2})^{\frac{1}{2}}} r\,dz\,d\theta\,dr$$
$$\int_0^2\int_0^{2\pi}\int_{r^2}^{({20-r^2})^{\frac{1}{2}}} r\,dz\,d\theta\,dr=2\pi\int_0^2r(({20-r^2})^{\frac{1}{2}}-{r^2})dr=28.2$$
$\endgroup$
