Evaluate the triple integral $\displaystyle \int\int\int_{E} (x + 2y) dV$, where $\displaystyle E$ is bounded by the surface $\displaystyle y = x^2$ and the planes $\displaystyle x = z$, $\displaystyle x = y$ and $\displaystyle z = 0$.

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- Apr 22nd 2009, 10:38 PMwik_chick88triple integral
Evaluate the triple integral $\displaystyle \int\int\int_{E} (x + 2y) dV$, where $\displaystyle E$ is bounded by the surface $\displaystyle y = x^2$ and the planes $\displaystyle x = z$, $\displaystyle x = y$ and $\displaystyle z = 0$.

- Apr 22nd 2009, 10:57 PMredsoxfan325
I'd like someone to verify this, but I'm pretty sure that this is it.

Getting the bounds is the hardest part. Just looking at $\displaystyle \mathbb{R}^2$, the area enclosed by $\displaystyle y=x^2$ and $\displaystyle y=x$ is the region such that $\displaystyle x\in[0,1]$. Note that here $\displaystyle x\geq x^2$.

Thus, we have

$\displaystyle x: 0\to1$

$\displaystyle y: x^2\to x$

$\displaystyle z: 0\to x$

Setting up the integral, we have $\displaystyle \int_0^1\int_{x^2}^x\int_0^x (x+2y)\,dz\,dy\,dx$

Here's my proposed solution if you want it.

__Spoiler__: