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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?

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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$$k\int_0^1\int_{-\sqrt{y}}^yx^2\left(y^5+1\right)dx\ dy=\frac{k}3\int_0^1\left(y^8+y^{13/2}+y^3+y^{3/2}\right) dy=\frac{483}{1620}k$$

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