Answer
The answer is below.
Work Step by Step
We find:
$(a)$
$\begin{aligned} \int_{0}^{1} \int_{x^{2}}^{\sqrt{x}}(x+y) d y d x &=\int_{0}^{1}\left[\frac{y^{2}}{2}+yx\right]_{x^{2}}^{\sqrt{x}} d x \\ &=\int_{0}^{1}\left(-\frac{x^{4}}{2}-x^{3}+\frac{x}{2}+x^{3 / 2}\right) d x \\ &=\left[\frac{2 x^{5 / 2}}{5}-\frac{x^{4}}{4}-\frac{x^{5}}{10}+\frac{x^{2}}{4}\right]_{0}^{1} \\ &=\frac{3}{10} \end{aligned}$
$(b)$
$\int_{-1}^{1} \int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}(y+x) d y d x=\int_{-1}^{1}\left[\frac{y^{2}}{2}+y x \right]_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}} d x$
$=\int_{-1}^{1}(2 x \sqrt{1-x^{2}}) d x$
$=\left[-\frac{2}{3}\left(-x^{2}+1\right)^{3 / 2}\right]_{-1}^{1}$
$=0$
