Find the volume of the solid bounded by $z=4x^2+4y^2, z=0, x^2+y^2=1$ and $x^2+y^2=2$.
What I know:
I know that when I draw the graph I will get two paraboloids giving me a radius of $1$ and $2$,$\theta= 2\pi$, and $z= 0$ to $4r^2$.
I keep getting a large number like $30\pi$ when the answer should $6\pi$.
Can anyone help me ? Thanks
Find the volume of the solid of revolution formed by revolving about the $x$ axis the figure in the $xz$-plane bounded by the graphs of $z=0$, $z=4x^2$, $x=1$ and $x=\sqrt{2}$.
$$ V=\int_1^{\sqrt{2}}2\pi rh\,dx$$
where $r=x$ and $h=4x^2$.
$$ V=2\pi\int_1^{\sqrt{2}}4x^3\,dx=2\pi[x^4]_1^{\sqrt{2}}=6\pi $$