• Support PF! Buy your school textbooks, materials and every day products Here!

Problem with polar double integral

  • Thread starter Xyius
  • Start date
  • #1
508
4
Hello everybody, I am having trouble doing this polar double integral. The problem says..

Find the area of the region..

[tex]\frac{1}{2}y^2 \leq x \leq 2y
0 \leq y \leq 8[/tex]

It is hard for me to come up with the limits of integration. Checking the answer would be easy because I can compare it to the rectangular integral which is easy, but I keep getting it wrong! The most trouble I am having is converting the line to a polar form and using it in the integral. I am pretty sure it would be theta = arctan(2) but I do not know where to put this in my integral.

For the upper limit on "r" I have [tex]4cot(\theta)csc(\theta)[/tex] I do not know what to put as the lower limit on r except the length r is at that point on the graph of [tex]4cot(\theta)csc(\theta)[/tex]. Which I got to be [tex]\frac{2}{sin(arctan(2)}[/tex]

I am very tired at the moment (Its 1:30 am!) and it is likely this doesn't make any sense to you. I just need some help. :\
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
250
Good morning Xyius! :zzz:
It is hard for me to come up with the limits of integration. Checking the answer would be easy because I can compare it to the rectangular integral which is easy, but I keep getting it wrong! The most trouble I am having is converting the line to a polar form and using it in the integral. I am pretty sure it would be theta = arctan(2) but I do not know where to put this in my integral.

For the upper limit on "r" I have [tex]4cot(\theta)csc(\theta)[/tex] I do not know what to put as the lower limit on r …
(btw, LaTeX here does not recognise line breaks … type "\ " for spaces :smile:)

The area is between a line and a parabola which intersect at the origin, so won't r go right down to zero? :wink:
 
  • #3
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,100
129
Moderator's note: thread moved from Calculus & Analysis

Hello everybody, I am having trouble doing this polar double integral. The problem says..

Find the area of the region..

[tex]\frac{1}{2}y^2 \leq x \leq 2y
0 \leq y \leq 8[/tex]
Are you sure about those limits for y? It's odd -- but not out of the question -- that those two curves will cross each other inside the limits of integration.

EDIT: Actually, in this case it is out of the question, since y=8 gives 32≤x≤16 for the limits on x, which is never true.
 
  • #4
508
4
Ah crap! I meant to write (1/4)y! It was late! :p

This is how I did my work, please tell me if this looks right to you.

So starting with the basic integral equation..
[tex]A=\int^{\phi_{2}}_{\phi_{1}}\int^{r_{2}}_{r_{1}}rdrd\phi[/tex]

First I want to find the "r" bounds so I did the following..

The graph suggests that my r will range from r=0 to the point where the line and the parabola meet. Since the line crosses through the origin, it acts like "r" and does not need to be taken into account. I will just simply start my r length at the point of intersection.

So I convert to polar coordinates for the parabola..

[tex]x=\frac{1}{4}y^2\rightarrow rcos\phi = \frac{1}{4}r^2sin^2\phi\rightarrow
rcos\phi - \frac{1}{4}r^2sin^2\phi \rightarrow r(cos\phi - \frac{1}{4}rsin^2\phi)
[/tex]

Therefore..
[tex]r=0,\\ and, \\ cos\phi - \frac{1}{4}rsin^2\phi = 0 \rightarrow
cos\phi = \frac{1}{4}rsin^2\phi \rightarrow \frac{4cos\phi}{sin^2\phi} = r
[/tex]

Therefore..
[tex]0\leq r \leq\frac{4cos\phi}{sin^2\phi} = 4cot\phi csc\phi[/tex]

So now that I have r out of the way, time to move on to theta.

Since the one equation is just a line crossing through the origin...

[tex]\phi =arctan( \frac{y}{x}) = arctan(2)[/tex]
(I just took any (x,y) pair on the line x=2y.

To get the next limit I set the expression for r equal to zero.
[tex]r = \frac{4cos\phi}{sin^2\phi}=0\rightarrow cos\phi = 0\rightarrow \phi=\frac{\pi}{2}[/tex]

So i get..
[tex]arctan(2) \leq \phi \leq \frac{\pi}{2}[/tex]


Plugging into the limits and evaluating I get..

[tex]A=\int^{\frac{\pi}{2}}_{arctan(2)}\int^{4cot\phi csc\phi}_{0}rdrd\phi \rightarrow

\frac{1}{2}\int^{\frac{\pi}{2}}_{arctan(2)}(16cot^2\phi csc^2\phi)d\phi,\ u=cot\phi, \

du=-csc^2\phi \rightarrow

8\int^{\frac{\pi}{2}}_{arctan(2)}u^2du \rightarrow \frac{8}{3}u^3 \rightarrow
[/tex]
[tex]
\frac{8}{3}cot^3\phi^{\frac{\pi}{2}}_{arctan(2)} \rightarrow

\frac{8}{3}[0-\frac{1}{8}] \rightarrow \frac{1}{3}

[/tex]

This doesn't make sense however because converting the integral into rectangular coordinates and doing it that way yeilds..

[tex]A=\frac{64}{3}[/tex]
 
  • #5
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,100
129
It all looks good, except for this:
Since the one equation is just a line crossing through the origin...

[tex]\phi =arctan( \frac{y}{x}) = arctan(2)[/tex]
(I just took any (x,y) pair on the line x=2y.
How about when y=1? What is x, and what is y/x in that case?
 
  • #6
508
4
A ha! It is 1/2 not 2! That makes it equivalent to the rectangular integral! Thanks a lot for all your guidance :)
 

Related Threads on Problem with polar double integral

Replies
1
Views
1K
Replies
4
Views
13K
Replies
16
Views
6K
Replies
3
Views
1K
Replies
13
Views
2K
Replies
13
Views
313
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
8
Views
2K
Top