A sculpture is given by the region bounded by the two surfaces in $(x, y, z)$ space defined by $z=(x^2 +y^2-1)^2$ and $z=4$. What is the volume of the sculpture?
I attempted to find the volume by setting up the triple integral,
$\iiint_{R} dx\space dy\space dz $
since $ z=(x^2+y^2-1)^2$,
finding the bounds for $dx$:
$1=(x^2+1-1)^2$
$x=1$
finding the bounds for $dy$:
$1=(x^2+y^2-1)^2$
$y=(2-x^2)^{\frac{1}{2}}$
hence the triple integral will take the form,
$ \int_{0}^{1} dx \space\int_{0}^{(2-x^2)^{1/2}} dy \space\int_{0}^{(x^2+y^2-1)^2}dz$
I tried to evaluate this intagral starting with $\int_{0}^{(x^2+y^2-1)^2}dz$ but it gave me a very lengthy integral with no cancellations, is this a correct way for going about the problem, or is there an easier approach to finding the volume?