A conversion of double integral to polar and evaluate, check

Question

Problem 1

Convert to polar form and solve

$$\int^{1}_{0}\int_{0}^{\sqrt{2y-y^2}}(1-x^2-y^2)\text{ dx dy}$$

$x^2+y^2-2y=0$

$x^2+(y-1)^2=1$

$x=rcos\theta$ $y=rsin\theta+1$

$r^2=1, r=1$

$$\int^{\pi}_{0}\int^{1}_{0}(1-r^2cos^2\theta-(y^2cos^2\theta+1)) rdrd\theta$$ $$\int^{\pi}_{0}\int^{1}_{0}(-r^3)drd\theta = -1/4{\pi}$$

Answer 1

You made an error when you substituted $y^2$. For the reference, the correct result is $\frac{2}{3} - \frac{\pi}{8}$.