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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.

Ivo Terek
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Anonymous
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    The natural way to set up the limits (to my mind) is to switch the inner limits on your first integral, since y^3 is less than y^2 for 0 < y < 1. Yes the second one gives the right value but then both iterations are running backwards from the larger number to the smaller one. – Ned Aug 14 '14 at 23:06
  • In the domain $0<y<1$, the curve $x=y^3$ lies below the curve $x=y^2$ so the limits of the $dx$ integral should be reversed from what you have. The 0 and 1 that you used are fine. – Paul Sundheim Aug 14 '14 at 23:06
  • As Ned and Paul both point out, your outer limits are fine; it is the $y^2$ and $y^3$ you want to switch since $x=y^3$ is to the left of $x=y^2$ for $0<y<1$. – user84413 Aug 14 '14 at 23:15

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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.

Derek Orr
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