Centroid of the region bounded by $z=x^2 + y^2 $, $z=0$, $-a<x<a$, and $-a<y<a$

Question

Since the given region is symmetric about the z-axis: $X_{c}=Y_{c}=0$

And,

$Z_{c}=( \int_{-a}^{a}\int_{-a}^{a} (x^2+y^2)^2 \, dx \, dy )/V = 14/15 a^2$

where,

$V=\int_{-a}^{a}\int_{-a}^{a} (x^2+y^2) \, dx \, dy = 8a^4/3$

I am pretty sure I evaluated the integrals correctly. However, the textbook says $Z_{c}=7/15a^2$. Why am I off by a factor of 2? Did I set up the equation for finding $Z_{c}$ correctly? Or is the textbook wrong?

Wolfram Alpha to confirm that I evaluated the integral correctly.

Answer 1

Figured it out:

$Z_{c}=( \int_{-a}^{a}\int_{-a}^{a} (x^2+y^2)^2 \, dx \, dy )/V = 14/15 a^2$

should be

$Z_{c}=( \int_{-a}^{a}\int_{-a}^{a} (x^2+y^2)^2/2 \, dx \, dy )/V = 7/15 a^2$

Because for every point on the surface $x^2+y^2$, the "centroid" of the line going up to that point is $(x^2+y^2)/2$