# Double Integral Question?

#### jk8985

##### New member
Find the double integral of (integral sign) (integral sign) ydA where D is the region bounded by (x+1)^2, x=y-y^3, x=-1, and y=-1

#### Jameson

Staff member
Hi jk8985,

Welcome to MHB!

Are you sure that the fourth part ($y=-1$) is correct? Here's what the first three equations look like when plotted and they have define a clear region but if you add $y=-1$ in there it becomes strange.

[GRAPH]itwfds8rx9[/GRAPH]

#### MarkFL

Staff member
Can you show us what you have tried so far? Have you determined what type of region $D$ is?

#### Chris L T521

##### Well-known member
Staff member
Restricting Jameson's graph of the region (including the graph of $y=-1$) appropriately gives us the region we're integrating over:

[graph]xk2g00psbd[/graph]

To evaluate the double integral over this region, you need to decide whether or not you should treat this as a Type I or Type II region (I'll just say that one way is much easier than the other).

Do you think you can determine the appropriate limits of integration and the double integral(s) needed to evaluate your original integral over this region?

#### Jameson

Staff member
Ah, yep I see my error now. Thanks for clearing that up Chris.

(I was looking in the wrong place for the region. $y=1$ will define a different region)

#### jk8985

##### New member
Restricting Jameson's graph of the region (including the graph of $y=-1$) appropriately gives us the region we're integrating over:

[graph]xk2g00psbd[/graph]

To evaluate the double integral over this region, you need to decide whether or not you should treat this as a Type I or Type II region (I'll just say that one way is much easier than the other).

Do you think you can determine the appropriate limits of integration and the double integral(s) needed to evaluate your original integral over this region?
I think it's a Type 2, but I'm not sure. I have no idea how to approach the limits of integration. Would you be able to provide an explanation on how to do so? It would be awesome

#### Jameson

Staff member
Start by trying to define the boundaries of the region. $x$ goes from what to what? $y$ goes from what to what?

#### jk8985

##### New member
Oh wait! I think I got it. Is this the integral?

∫0 to -1 ∫(-1 to y−y^3) y dxdy

plus

∫0 to 1 ∫(√(y−1)) to (y−y^3) ydxdy

I don't really understand why these are the limits of integration though, and why they are added together.

#### MarkFL

Staff member
I would look first at the region below the $x$-axis, we'll call it $D_1$. This is in fact a type II region. Now since $D_1$ is type II, we may write:

$$\displaystyle \underset{D_1}\iint y\,dA=\int_c^a\!\int_{h_1(y)}^{h_2(y)} y\,dx\,dy$$

Can you determine the limits now?

Okay, I see you have responded as I am composing this post. Your first integral is nearly correct, you have the outer limits reversed. And for the second integral your inner limits contain an error where you solved the quadratic for $x$.

Can you restate the two integrals now?

#### jk8985

##### New member
∫-1(lower limit) to 0 (upper limit) ∫(-1 (lower limit) to y−y^3 (upper limit)) y dxdy

plus

∫0 (lower limit) to 1 (upper limit) ∫(√(y)-1)) (lower limit) to (y−y^3) (upper limit) y dxdy

I did mine as two different double integrals. Is that okay?
Did I get the two sets of double integrals correct this time?

#### MarkFL

Staff member
∫-1(lower limit) to 0 (upper limit) ∫(-1 (lower limit) to y−y^3 (upper limit)) y dxdy

plus

∫0 (lower limit) to 1 (upper limit) ∫(√(y)-1)) (lower limit) to (y−y^3) (upper limit) y dxdy

I did mine as two different double integrals. Is that okay?
Did I get the two sets of double integrals correct this time?
Yes, that's correct:

$$\displaystyle \underset{D}\iint y\,dA=\underset{D_1}\iint y\,dA+\underset{D_2}\iint y\,dA=\int_{-1}^{0}\!\int_{-1}^{y-y^3} y\,dx\,dy+\int_{0}^{-1}\!\int_{\sqrt{y}-1}^{y-y^3} y\,dx\,dy$$

We may split it up this way since $$\displaystyle D=D_1\,\cup\,D_2$$.

Can you now evaluate the iterated integrals?

#### jk8985

##### New member
I get a non-real result when doing from 0 to -1 for the second double integral. for the first set of double integrals i get 11/30

#### MarkFL

Staff member
I get a non-real result when doing from 0 to -1 for the second double integral. for the first set of double integrals i get 11/30
My gravest apologies, the outer upper limit should be $1$...I made a silly typo there.

Let's have a look see...

i) $D_1$:

$$\displaystyle \int_{-1}^{0}\!\int_{-1}^{y-y^3} y\,dx\,dy=\int_{-1}^{0}\left(y\left(y-y^3-(-1) \right) \right)\,dy=\int_{-1}^{0}-y^4+y^2+y\,dy=$$

$$\displaystyle \left[-\frac{1}{5}y^{5}+\frac{1}{3}y^3+\frac{1}{2}y^2 \right]_{-1}^0=0-\left(\frac{1}{5}-\frac{1}{3}+\frac{1}{2} \right)=-\frac{11}{30}$$

i) $D_2$:

$$\displaystyle \int_{0}^{1}\!\int_{\sqrt{y}-1}^{y-y^3} y\,dx\,dy=\int_0^1\left(y\left(y-y^3-\left(\sqrt{y}-1 \right) \right) \right)\,dy=\int_0^1 -y^4+y^2-y^{\frac{3}{2}}+y\,dy=$$

$$\displaystyle \left[-\frac{1}{5}y^5+\frac{1}{3}y^3-\frac{2}{5}y^{\frac{5}{2}}+\frac{1}{2}y^2 \right]_0^1=\left(-\frac{1}{5}+\frac{1}{3}-\frac{2}{5}+\frac{1}{2} \right)-0=\frac{7}{30}$$

I have checked these result with a CAS, and they agree.