Problem: I am trying to calculate the mass for the solid bounded by $z = y^5 + 1, \ z = 0, \ y = x, \ y = x^2, \ y = 1.$ The density at each point is directly proportional to the square of the distance from the yz-plane.
I split the integral to simplify the calculation:
Assuming $\rho=kx^2$,
$k \cdot \int_{-1}^0 \int_{x^2}^1 \int_0^{y^5 + 1} x^2 dz \ dy \ dx + k \cdot \int_{0}^1 \int_x^{1} \int_0^{y^5 + 1} x^2 dz \ dy \ dx$
I am wondering if there is an easier way to set up the triple integral and if my limits of integration are correct?
Your limits of integration are perfectly correct, but there is an easier way to set up that integral. If you let $y$ go from $0$ to $1$ and $x$ go from $-\sqrt{y}$ to $y$, you can express it as a single integral: $$k\int_0^1\int_{-\sqrt{y}}^y\int_0^{y^5+1}x^2\ dz\ dx\ dy$$ Both this and your formulation give the same answer $(\approx 0.298k)$, which I've shown here along with a visual representation of the region of integration in the $xy$-plane.
Edit: The exact answer is given by$$k\int_0^1\int_{-\sqrt{y}}^yx^2\left(y^5+1\right)dx\ dy=\frac{k}3\int_0^1\left(y^8+y^{13/2}+y^3+y^{3/2}\right) dy=\frac{483}{1620}k$$