Problem with polar double integral

In summary: This is how I did my work, please tell me if this looks right to you.In summary, the person is having trouble integrating a polar integral and is having trouble coming up with the limits for y. They convert the line to polar coordinates and find that the upper limit for r is 4cot(\theta)csc(\theta). They also find the lower limit for r to be the length r at the point where the line and the parabola intersect.
  • #1
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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. :\
 
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  • #2
Good morning Xyius! :zzz:
Xyius said:
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
Moderator's note: thread moved from Calculus & Analysis

Xyius said:
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
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
It all looks good, except for this:
Xyius said:
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
A ha! It is 1/2 not 2! That makes it equivalent to the rectangular integral! Thanks a lot for all your guidance :)
 

1. What is a polar double integral?

A polar double integral is a type of integral used to find the area under a polar curve in a two-dimensional coordinate system. It is similar to a regular double integral, but the integration is done in polar coordinates rather than Cartesian coordinates.

2. How is a polar double integral different from a regular double integral?

A polar double integral uses polar coordinates, which are represented by a radius and an angle, instead of Cartesian coordinates, which are represented by x- and y-coordinates. This means that the limits of integration and the integrand will be expressed in terms of polar coordinates.

3. What are the challenges of solving a polar double integral?

The main challenge in solving a polar double integral is converting the integrand from Cartesian coordinates to polar coordinates. This can be done using trigonometric identities and substitution. Additionally, the limits of integration may also need to be converted to polar coordinates, which can be tricky for complex curves.

4. How can I determine the limits of integration for a polar double integral?

The limits of integration for a polar double integral depend on the shape and orientation of the polar curve. Generally, the radius will have a lower limit of 0 and an upper limit of the distance from the origin to the curve. The angle will have a lower limit of 0 and an upper limit of 2π, or the full rotation around the origin. However, in some cases, the limits may need to be adjusted to only cover a portion of the curve.

5. What are some real-world applications of polar double integrals?

Polar double integrals have many applications in physics, engineering, and other fields. They can be used to calculate the mass, center of mass, and moment of inertia of objects with curved shapes. They are also used in calculating the electric field and gravitational potential of objects with circular symmetry. In addition, polar double integrals are commonly used in calculating volumes and surface areas of three-dimensional shapes with curved surfaces, such as spheres or cones.

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