I was given this problem:
Find an integral equal to the volume of the solid bounded by $z=4-2y,z=0,x=y^4,x=1$ and evaluate.
I understand how to evaluate once my double integral is set up, but I do not know how to find my limits of integration.
I am assuming that my function will be $z=4-2y$ and that using this I should be able to find my limits of integration. I can say that $0=4-2y$ which means that $y=2$. I can then plug that into $x=y^4$ and get $1\leq x\leq 16$ which may be correct, but I still am missing the limits of integration for y.
Am I thinking about this problem correctly? How can I go about solving this?
Best Answer
Note that $x=y^4$ and $x=1$ intersect at $ (x,y)=(1,\pm1)$. which define the limits for the integration region in the $xy$- plane. Thus, the volume integral is
$$\int_{-1}^1 \int_{y^4}^1 (4-2y)dxdy =\frac{32}5$$