How Do You Calculate the Double Integral Over a Complex Region?

In summary, the double integrals are: $∫-1(lower limit) to 0 (upper limit) ∫(-1 (lower limit) to y−y^3 (upper limit)) y dxdy$ and $∫0 (lower limit) to 1 (upper limit) ∫(√(y)-1)) (lower limit) to (y−y^3) (upper limit) y dxdy$.
  • #1
jk8985
12
0
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
 
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  • #2
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]
 
  • #3
Can you show us what you have tried so far? Have you determined what type of region $D$ is?
 
  • #4
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)
 
  • #5
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)
 
  • #6
Chris L T521 said:
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
 
  • #7
Start by trying to define the boundaries of the region. $x$ goes from what to what? $y$ goes from what to what?
 
  • #8
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.
 
  • #9
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?
 
  • #10
∫-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?
 
  • #11
jk8985 said:
∫-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?
 
  • #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
 
  • #13
jk8985 said:
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.
 
  • #14
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
 
  • #15
jk8985 said:
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)

We do ask that you refrain from drawing attention in one thread to other threads you have posted. Doing this can make the original thread unnecessarily convoluted, and it is redundant as well. :D

We ask that once you post a thread to be patient and wait for someone who has solid help to offer to do so. All new threads show up in the search for new posts, and so those of us offering help will see them.
 

FAQ: How Do You Calculate the Double Integral Over a Complex Region?

What is a double integral?

A double integral is a mathematical concept that involves calculating the signed area between a two-dimensional region and a plane. It is represented by two nested integrals and is used to solve problems related to volume, mass, and moments of inertia.

How do you solve a double integral?

To solve a double integral, you first need to identify the limits of integration for both the inner and outer integrals. Then, you need to evaluate the inner integral and substitute the result into the outer integral. Finally, you evaluate the outer integral to get the final answer.

What is the difference between a single and double integral?

A single integral calculates the area under a curve in one dimension, whereas a double integral calculates the signed area between a two-dimensional region and a plane. In other words, a single integral deals with one variable, whereas a double integral deals with two variables.

What are the applications of double integrals?

Double integrals have various applications in physics, engineering, economics, and other fields. Some common applications include calculating volumes of irregularly shaped objects, finding the center of mass of a two-dimensional object, and determining the total mass of an object with varying density.

Is there a specific order to evaluate a double integral?

Yes, there is a specific order to evaluate a double integral, known as the "order of integration." The inner integral is always evaluated first, and then the result is substituted into the outer integral. This order can be reversed, depending on the problem, but the inner integral is always evaluated first.

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