Double Integral Confusion: How Do I Handle Boundaries with U-Substitution?

In summary, to find ʃʃ cos(x+2y)dA you need to integrate dx first and dy second, and if that's your choice of order you will need a u-substitution for each.
  • #1
magnifik
360
0
i am confused about the double integral ʃʃ cos(x+2y)dA, where R = [0,pi]x[0,pi/2]
i realize for the integral that i must do u-substitution. when i do this, however, do i also have to change the boundary conditions as in a single integral?

i got -8 without changing the boundary conditions, but I'm not sure if that's right.
i will show my work if anyone wants to check how i got there
 
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  • #2
The u-substitutions you have to do are pretty easy. But I don't get -8. Maybe you'd better show how you did it.
 
  • #3
magnifik said:
i am confused about the double integral ʃʃ cos(x+2y)dA, where R = [0,pi]x[0,pi/2]
i realize for the integral that i must do u-substitution. when i do this, however, do i also have to change the boundary conditions as in a single integral?

i got -8 without changing the boundary conditions, but I'm not sure if that's right.
i will show my work if anyone wants to check how i got there

I did this out quickly using [tex] \cos u ={{{\rm e}^{{\rm j} u} + {\rm e}^{-{\rm j} u}}\over{2}}[/tex], and didn't get -8 either. It's probably just a simple error in the substitution process.
 
  • #4
stevenb said:
I did this out quickly using [tex] \cos u ={{{\rm e}^{{\rm j} u} + {\rm e}^{-{\rm j} u}}\over{2}}[/tex], and didn't get -8 either. It's probably just a simple error in the substitution process.

Why not just use the sum formula for cosine?

[tex] cos(A + B) = cosAcosB - sinAsinB. [/tex]
 
  • #5
Raskolnikov said:
Why not just use the sum formula for cosine?

[tex] cos(A + B) = cosAcosB - sinAsinB. [/tex]

Just used the first thought that came to my mind. It is so simple by either method, and I didn't bother to consider all methods to find the best. I just wanted to get a number to help the OP know if he was right or wrong. Actually, Dick beat me too it, so I just provided a second verification.
 
  • #6
hm, it seems no one answered my question: do i have to change the boundary conditions when doing the u-substitution?
 
  • #7
yes, you do. out of curiosity, which u substitution did you use?
 
  • #8
magnifik said:
hm, it seems no one answered my question: do i have to change the boundary conditions when doing the u-substitution?

Integrate dx first and dy second, if that's your choice of order. You'll need a u-substitution for each, and yes, you may need a change of limits. Depending on how you do it. It's kind of useless to debate this until you show how you arrived at the wrong answer.
 
  • #9
using the sum formula for cosine i got -2?
 
Last edited:
  • #10
magnifik said:
using the sum formula for cosine i got -2?

That looks correct
 
  • #11
on a somewhat unrelated note, is ln2 the correct answer for ʃʃ (xe^x)/y dydx for R = [0,1]X[1,2]? i did integration by parts for the xe^x part but had the same issue of whether or not to change the start/end point.
 
  • #12
magnifik said:
on a somewhat unrelated note, is ln2 the correct answer for ʃʃ (xe^x)/y dydx for R = [0,1]X[1,2]? i did integration by parts for the xe^x part but had the same issue of whether or not to change the start/end point.

Looks correct to me.
 
  • #13
magnifik said:
... but had the same issue of whether or not to change the start/end point.

Yes, this seems to be a question in your mind. The simple rule is that you only need to change limits if you have a substitution that results in a change of variables. If your functions are still using x and y, and your integration is still over dx and dy, then there is no need to think about changing limits. However, if you change variables, such as u=2x, and/or w=sin(y), then you will generally need to change the limits.
 
  • #14
stevenb said:
Yes, this seems to be a question in your mind. The simple rule is that you only need to change limits if you have a substitution that results in a change of variables. If your functions are still using x and y, and your integration is still over dx and dy, then there is no need to think about changing limits. However, if you change variables, such as u=2x, and/or w=sin(y), then you will generally need to change the limits.

thank you
 

1. What is a double integral and why is it confusing?

A double integral is a type of integral in calculus that involves integrating a function of two variables over a region in a two-dimensional plane. It is often confusing because it involves multiple steps and can be difficult to visualize.

2. How is a double integral different from a single integral?

A single integral involves integrating a function of one variable over a one-dimensional interval, whereas a double integral involves integrating a function of two variables over a two-dimensional region.

3. What is the purpose of a double integral in mathematics?

A double integral is used to calculate the volume of a solid in three-dimensional space, as well as to find the surface area of a curved surface. It is also used in many applications in physics and engineering.

4. What are some common mistakes made when solving double integrals?

Some common mistakes include forgetting to include the differential (dx or dy) when setting up the integral, mixing up the order of integration, and not properly defining the limits of integration.

5. How can I improve my understanding of double integrals?

Practice, practice, practice! It is important to understand the concept and steps involved in solving a double integral, but the best way to improve is to work through different examples and practice problems. You can also seek help from a tutor or teacher if you are still struggling to grasp the concept.

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