triple integral

• Apr 22nd 2009, 10:38 PM
wik_chick88
triple 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 PM
redsoxfan325
Quote:

Originally Posted by wik_chick88
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 $\displaystyle z = 0$.

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:

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

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

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

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

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

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