1) \(\int\limits_0^a\int\limits_x^a(x^2+y^2)dydx\)
Region is bounded from y = x to y = a and x = 0 to x = a.
Now, \(\int\limits_0^a\int\limits_x^a(x^2+y^2)dydx\) = \(\int\limits_0^a[x^2y+\frac{y^3}3]_x^adx\)
= \(\int\limits_0^a(ax^2+\frac{a^3}3-\frac{x^3}3-x^3)dx\)
= \(\int\limits_0^a(-\frac43x^3+ax^2+\frac{a^3}3)dx\)
= \(-\frac43[\frac{x^4}4]_0^a+a[\frac{x^3}3]_0^a+\frac{a^3}3[x]_0^a\)
= \(-\frac13(a^4-0)+\frac a3(a^3-0)+\frac{a^3}3(a-0)\)
= a4(\(-\frac13+\frac13+\frac13\))
= \(\frac{a^4}3\)
2)
\(\int\limits_{x=0}^{x=a}\left(\int\limits_{y=0}^{y=\frac ba\sqrt{a^2-x^2}}ydy\right)dx\)
= \(\int\limits_0^ax[\frac{y^2}2]_0^{\frac ba\sqrt{a^2-x^2}}dx\)
= \(\int\limits_0^a\frac{x}2((b/a)^2(a^2-x^2)-0)dx\)
= \(\frac{b^2}{2a^2}\int\limits_0^a(a^2x-x^3)dx\)
= \(\frac{b^2}{2a^2}[\frac{a^2x^2}2-\frac{x^4}4]_0^a\)
= \(\frac{b^2}{2a^2}(\frac{a^4}2-\frac{a^4}4)\)
= \(\frac{b^2}{2a^2}\times\frac{a^4}4\) = \(\frac{a^2b^2}8\)
