Find the volume under the surface $z=2x+y^2$ above the region bounded by $x=y^2$ and $x=y^3$.
I have already worked out the solution to the problem, but I used $\int_0^1\int_{y^2}^{y^3} 2x+y^2 \newcommand{\d}{\mathrm{d}} \d x \d y$ instead of $\int_1^0\int_{y^2}^{y^3} 2x+y^2 \d x \d y$ and hence got the negative of the correct answer.
This may be inane, but can someone tell me why the limits of $y$ must be $0 \to 1$ and not the other way around? My mind isn't what it used to be. Thanks in advance.
Best Answer
Since $y^3$ is the higher term and $y^2$ is the lower term in the first integral, you'll get (essentially) $y^3 - y^2$. However, $y^2 > y^3$ between $0$ and $1$. So you have to change the order and put $1$ on the bottom and $0$ on top.