Double Integral problem What am I suppose to do? Related to polar coordinates.

In summary, the conversation is about a problem with an image shown and the speaker's confusion about the limits \theta ranging from 0 to cos(\theta). They mention getting an incorrect answer of 3/16 pi^2 and ask for guidance. Another person suggests drawing the lines y=0 and y=x to visualize the problem, and the speaker expresses gratitude for the help.
  • #1
DavidAp
44
0
The problem and my work is shown in the image below. However, I feel like I did something horrible wrong but I'm not sure where!

24xjrl5.jpg


I'm sorry if my handwriting is illegible. If you're having difficulties please leave a comment and I will not hesitate to type it out as a response. Any guidance is greatly appreciated.

Thank you for taking the time to review my question.
 
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  • #2
You've got [itex]\theta[/itex] ranging from [itex]0[/itex] to [itex]cos(\theta)[/itex], that doesn't make any sense. I'm fairly sure you should have [itex]\theta[/itex] going from [itex]0[/itex] to [itex]\frac{\pi}{4}[/itex]. Which you'll be able to see if you draw a picture of [itex]R[/itex].
 
  • #3
Stimpon said:
You've got [itex]\theta[/itex] ranging from [itex]0[/itex] to [itex]cos(\theta)[/itex], that doesn't make any sense. I'm fairly sure you should have [itex]\theta[/itex] going from [itex]0[/itex] to [itex]\frac{\pi}{4}[/itex]. Which you'll be able to see if you draw a picture of [itex]R[/itex].

I don't know how to draw 0 <= y <= x. I think that might be part of my problem...

I did something wrong though when using pi/2! The answer is 3/64 pi^2 but I keep getting 3/16 pi^2! Why is my denominator 4x less than the answer? Here's my work.

∫(1 -> 2) ∫(0 -> pi/2) θr drdθ
1/2 ∫(1 -> 2) θ(r^2)(0 -> pi/2) dθ
1/2 ∫(1 -> 2) θ(pi^2)/4 dθ
(pi^2)/8 ∫(1 -> 2) θ dθ
(pi^2)/16 (θ^2)(1 -> 2)
(pi^2)/16 (2^2 - 1^1)
(pi^2)/16 (4 - 1)
(pi^2)/16 (3)
3/16 pi^2

Any insight on what I did wrong?
 
  • #4
Ah I think I didn't edit my post in time then, the upper limit should have been [itex]\frac{\pi}{4}[/itex], sorry. Your method is fine.

And just draw the lines [itex]y=0[/itex] and [itex]y=x[/itex] and you should be able to see where [itex]y[/itex] satisfies [itex]0{\leq}y{\leq}x[/itex]. Or think about it this way, where does [itex](x,y)[/itex] satisfy [itex]0{\leq}y[/itex] and where does it satisfy [itex]y{\leq}x[/itex]. Then you just find the intersection of those two areas.
 
  • #5
Stimpon said:
Ah I think I didn't edit my post in time then, the upper limit should have been [itex]\frac{\pi}{4}[/itex], sorry. Your method is fine.

And just draw the lines [itex]y=0[/itex] and [itex]y=x[/itex] and you should be able to see where [itex]y[/itex] satisfies [itex]0{\leq}y{\leq}x[/itex]. Or think about it this way, where does [itex](x,y)[/itex] satisfy [itex]0{\leq}y[/itex] and where does it satisfy [itex]y{\leq}x[/itex]. Then you just find the intersection of those two areas.
Thank you so much, not just on this problem but on future problem to come! I never thought of visualizing it that way!
 

1. What is a double integral problem in polar coordinates?

A double integral problem in polar coordinates is a mathematical problem that involves finding the double integral of a function over a region in the polar coordinate plane. This can be thought of as finding the volume under a surface in three-dimensional space.

2. How do I set up a double integral in polar coordinates?

To set up a double integral in polar coordinates, you need to express the function in terms of polar coordinates, then determine the limits of integration for r and theta. The limits of integration for r will depend on the shape of the region, while the limits for theta will typically be 0 to 2π.

3. What is the difference between a single integral and a double integral in polar coordinates?

A single integral in polar coordinates involves finding the area under a curve in the polar coordinate plane, while a double integral involves finding the volume under a surface in three-dimensional space. Double integrals in polar coordinates are more complex than single integrals because they involve integrating over both r and theta.

4. How do I know when to use a double integral in polar coordinates?

A double integral in polar coordinates is typically used when the region of integration is better described in terms of polar coordinates, such as when the region has circular or radial symmetry. It can also be useful when the function being integrated is more easily expressed in polar coordinates.

5. What are some common mistakes to avoid when solving double integral problems in polar coordinates?

Some common mistakes to avoid when solving double integral problems in polar coordinates include forgetting to convert the function to polar coordinates, using the wrong limits of integration, and forgetting to include the Jacobian factor (r) when setting up the integral. It is also important to carefully consider the shape of the region and choose the appropriate order of integration.

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