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- Thread starter jk8985
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- Jan 26, 2012

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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]

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- Jan 26, 2012

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[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?

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- Jan 26, 2012

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

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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

[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?

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- Jan 26, 2012

- 4,055

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

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?

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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?

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Yes, that's correct:

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?

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

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

Can you now evaluate the iterated integrals?

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My gravest apologies, the outer upper limit should be $1$...I made a silly typo there.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

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.

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

http://mathhelpboards.com/calculus-10/angle-between-two-planes-8180.html

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Glad to hear it!

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

http://mathhelpboards.com/calculus-10/angle-between-two-planes-8180.html

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