# Double integrals over general regions

Set up iterated integrals for both orders of integration. Then evaluate the double integral using the easier order and explain why it's easier. $$\int\int_{D}{ydA}, \text{D is bounded by y=x-2, x=y^2}$$
I'm having trouble with setting up the integral for both type I and type 2 and choosing which one would be easier. I'm sure I'll be able to integrate it with no problem, but setting them up is where I would need some help.

If you draw region $D$ you will have this:

To find cutoffs you solve the system

$x=y+2$ and $x=y^2$

so

$y^2-y-2=0$

then the points are $(1,-1)$ and $(4,2)$.

Now, to setting up the integral over type $1$ region we need to partition $D$ in two subregions, so

$$\int {\int_D {ydA = \int_0^1 {\int_{ - \sqrt x }^{\sqrt x } {ydydx + } } } } \int_1^4 {\int_{x - 2}^{\sqrt x } {ydydx} }$$

To setting up the integral over type $2$ region

$$\int_{ - 1}^2 {\int_{{y^2}}^{y + 2} {ydxdy} }$$

It seems that second integral is easier than the first one.

• I would accept this answer, it is explained very clearly. – Sebastialonso Jul 26 '13 at 22:52

Plotting both $y=x-2$ and $x=y^2$ in wolfram alpha:

Based upon the points of intersection, you can see that the region D exists for $(x,y)$ with $x\in[1,4]$ and $y\in[-1,2]$.

• Yeah, I graphed it on my calculator, and I know the intersection points, but I'm still having trouble setting up the integral though. – Jc E Jul 24 '13 at 20:52

I found the answer. $\int_{0}^1\int_{-\sqrt{x}}^\sqrt{x}{ydydx}+\int_{1}^4\int_{x-2}^\sqrt{x}{ydydx}=\int_{-1}^2\int_{y^2}^{x+2}{ydxdy}$
It's easier to to integrate $\int_{-1}^2\int_{y^2}^{x+2}{ydxdy}$ and after integrating it, the answer comes out to be $\frac94$

• That's indeed correct! – Etienne Jul 24 '13 at 21:36

A type 2 region will work. Observe the graph if done as a type 2:

$$\int_a^b\int_{h_1(y)}^{h_2(y)}y\,dx\,dy$$

First you need to find where the two intersect:

$$y+2=y^2$$

$$y^2-y-2=0$$

$$y=-1;y=2$$

$$(1,-1),(4,2)$$

It is clear that the bounds are as follows:

\begin{align} h_1(y)&=y^2 \\ h_2(y)&=y+2 \\ a&=-1 \\ b&=2 \end{align}

• The answer in the back of the book shows that it's both type I and type II – Jc E Jul 24 '13 at 20:55
• @Jc-E But you asked which one would be easier to solve, which is the type 2 region ZettaSuro showed. One can easily see from the region drawn on this answer that the integral becomes $\int^2_{-1}\int^{y+2}_{y^2} y dxdy$ – Alex Jul 24 '13 at 21:21
• @Alex I had previously said that the only way to do it was with a type I region, but I edited afterwards to say that it's easier. The edit doesn't show up because I edited within 5 minutes of posting. – Ataraxia Jul 24 '13 at 21:23