Set up the limits and evaluate $\iiint ydv$ Set up the limits and evaluate the integral.
$$\displaystyle\iiint ydv$$
$$G$$
G is the region enclosed by the plane $z = y$, $xy-plane$ and the surface $y = 1 - x^2$
I need help finding the limits of integration. I have this so far, but I don't think it is correct.
$$\displaystyle\int_{0}^{1}\int_{0}^{1 - x^2}\int_{0}^{y} y\;dz\;dy\;dx$$
 A: $\displaystyle\int_{0}^{1}\int_{0}^{1 - x^2}\int_{0}^{y} y\;dz\;dy\;dx$
Taking the antiderivative in terms of $dz$, we have
$\displaystyle\int_{0}^{1}\int_{0}^{1 - x^2}\int_{0}^{y} yz\;dy\;dx$
Now, evaluate from $F(y) -f(0)$
$\displaystyle\int_{0}^{1}\int_{0}^{1 - x^2}\int_{0}^{y} yy\;dy\;dx$
$\displaystyle\int_{0}^{1}\int_{0}^{1 - x^2}\int_{0}^{y} y^2\;dy\;dx$
Taking the antiderivative in terms of $dy$, we have
$\displaystyle\int_{0}^{1}\int_{0}^{1 - x^2}\int_{0}^{y} \frac{y^3}{3}\;dx$
Now, evaluate from $F(1-x^2) -f(0)$
$\displaystyle\int_{0}^{1}\int_{0}^{1 - x^2}\int_{0}^{y} \frac{(1-x^2)^3}{3}\;dx$
We need to expand first 
$\displaystyle\int_{0}^{1}\int_{0}^{1 - x^2}\int_{0}^{y} \frac{-x^6+3x^4-3x^2+1}{3}\;dx$
and simplify 
$\displaystyle\int_{0}^{1}\int_{0}^{1 - x^2}\int_{0}^{y} \frac{-x^6}{3}+x^4-x^2+\frac{1}{3}\;dx$
Taking the antiderivative in terms of $dx$, we have
$\displaystyle\int_{0}^{1}\int_{0}^{1 - x^2}\int_{0}^{y} \frac{-x^7}{21}+\frac{x^5}{5}-\frac{x^3}{3}+\frac{x}{3}$
Evaluate from $f(1) -f(0)$
$\displaystyle\int_{0}^{1}\int_{0}^{1 - x^2}\int_{0}^{y} \frac{-1}{21}+\frac{1}{5}-\frac{1}{3}+\frac{1}{3}$
The final answer is $\frac{16}{105}$
