triple integral

• Apr 22nd 2009, 11:38 PM
wik_chick88
triple integral
Evaluate the triple integral $\int\int\int_{E} (x + 2y) dV$, where $E$ is bounded by the surface $y = x^2$ and the planes $x = z$, $x = y$ and $z = 0$.
• Apr 22nd 2009, 11:57 PM
redsoxfan325
Quote:

Originally Posted by wik_chick88
Evaluate the triple integral $\int\int\int_{E} (x + 2y) dV$, where $E$ is bounded by the surface $y = x^2$ and the planes $x = z$, $x = y$ and $z = 0$.

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 $\mathbb{R}^2$, the area enclosed by $y=x^2$ and $y=x$ is the region such that $x\in[0,1]$. Note that here $x\geq x^2$.

Thus, we have

$x: 0\to1$
$y: x^2\to x$
$z: 0\to x$

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

Here's my proposed solution if you want it.

Spoiler:

Evaluating the innermost integral gives us $(x+2y)z\big|_0^x = x^2+2xy$.

Now the integral is $\int_0^1\int_{x^2}^x(x^2+2xy)\,dy\,dx$

Evaluating the next part of the integral gives us $(x^2y+xy^2)\big|_{x^2}^x = x^3+x^3-x^4-x^5 = 2x^3-x^4-x^5$

Now the integral is $\int_0^1 (2x^3-x^4-x^5)\,dx$

Evaluating this gives us $\left(\frac{x^4}{2}-\frac{x^5}{5}-\frac{x^6}{6}\right)\bigg|_0^1 = \frac{1}{2}-\frac{1}{5}-\frac{1}{6} = \frac{2}{15}$

So I think $\frac{2}{15}$ is the answer.