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Double Integral Question?

jk8985

New member
Dec 16, 2013
12
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

Administrator
Staff member
Jan 26, 2012
4,043
Hi jk8985, (Wave)

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

Administrator
Staff member
Feb 24, 2012
13,775
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
Jan 26, 2012
995
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? (Smile)
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
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
Dec 16, 2013
12
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? (Smile)
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 :D
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
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
Dec 16, 2013
12
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Dec 16, 2013
12
∫-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

Administrator
Staff member
Feb 24, 2012
13,775
∫-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
Dec 16, 2013
12
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

Administrator
Staff member
Feb 24, 2012
13,775
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.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
awesome, exactly what I got when I did it.

If you could help me with this, it would be awesome :)

http://mathhelpboards.com/calculus-10/angle-between-two-planes-8180.html
Glad to hear it! (Yes)

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