Question

integrate x(y^3+1)^(1/2) dydx
over (x/3)<=y<=2 and 0<=x<=6

Answer 1

First draw your region D.


So you want to integrate x first. To do this you need to switch the bounds. So now we have:
\[0 \le x \le 3y; 0 \le y \le 2\]
Giving:
\[\int\limits_0^2 \int\limits_0^{3y}x(y^3+1)^{\frac{1}{2}}dx dy=\frac{1}{2}\int\limits_0^2 (3y)^2(y^3+1)^{\frac{1}{2}}dy=\frac{9}{2}\int\limits_0^1y^2(y^3+1)^{\frac{1}{2}}dy\]
Now make a change of variable on y^3+1. Now it's easy.