Converting into polar coordinates for integration

In summary, the given double integral can be converted into polar coordinates by changing the limits of integration to r = 0 to 1 and \theta = 0 to 2\pi, as the region of integration is the unit circle. Also, the area element for a polar integral is dA = rdrd\theta. The final integral in polar coordinates is \int_{0}^{2\pi} \int_{0}^{1} r\;ln\left(r^2 + 1\right) dr\;d\theta. It is important to draw a graph of the region of integration to better understand the limits of integration.
  • #1
VinnyCee
489
0
Another problem that I cannot figure out. Convert the follorwing into polar coordinates:

[tex]\int_{-1}^{1} \int_{-\sqrt{1 - y^2}}^{\sqrt{1 - y^2}} ln\left(x^2 + y^2 + 1\right) dx\;dy[/tex]

I did this so far:

[tex]ln\left(x^2 + y^2 + 1\right) = ln\left(r^2 + 1\right)[/tex]

[tex]\sqrt{1 - y^2} = \sqrt{1 - r^2 \sin^2 \theta}[/tex]

Now what do I do?

Is this possibly right?

[tex]\int_{0}^{2\pi} \int_{-\sqrt{1 - r^2 \sin^2 \theta}}^{\sqrt{1 - r^2 \sin^2 \theta}} ln\left(r^2 + 1\right) dr\;d\theta[/tex]
 
Last edited:
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  • #2
Trig identities, what does [itex] 1-sin^2(\theta) [/tex] equal?
 
  • #3
Better yet,

x^2+y^2 = r^2
with r = 1,
x^2+y^2 = 1
x^2 = 1-y^2

x = sqrt(1-y^2)
 
  • #4
If what whozum said isn't clear to you, try drawing a graph of the region you're integrating over. Also, remember that the area element for a polar integral is

[tex]dA=rdrd\theta[/tex]
 
  • #5
Is this it?

Including the [tex]dA = r\;dr\;d\theta[/tex], is this integral the right conversion?

[tex]\int_{0}^{2\pi} \int_{-\sqrt{1 - r^2 \sin^2 \theta}}^{\sqrt{1 - r^2 \sin^2 \theta}} r\;ln\left(r^2 + 1\right) dr\;d\theta[/tex]
 
  • #6
The limits of r simplify, other than that your right. Look at our hints above.
 
  • #7
Cool , but how does the [tex]1-sin^2(\theta) = \cos^2\left(\theta\right)[/tex] help when I have these limits of integration for [itex]r[/itex]?

[tex]\sqrt{1 - r^2 \sin^2 \theta}}[/tex]

It does not factor right?

Attached is a graph.
 

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  • #8
VinnyCee said:
[tex]\int_{0}^{2\pi} \int_{-\sqrt{1 - r^2 \sin^2 \theta}}^{\sqrt{1 - r^2 \sin^2 \theta}} r\;ln\left(r^2 + 1\right) dr\;d\theta[/tex]

Does it make sense to have "r" in the limit of an integral over r?
 
  • #9
It does factor, that's my point. Read what I said above clearly, sketch the domain. It will be clear what your limits for R are.
 
  • #10
Let's try this?

[tex]x = \sqrt{1 - y^2}[/tex]
[tex]r\;\cos\theta = \sqrt{1 - r^2\;\sin^2\theta}[/tex]

Then solve for r?
 
  • #11
Like this?

[tex]r^2\;\cos^2\theta = 1 - r^2\;\sin^2\theta[/tex]

[tex]r^2\;\left(\cos^2\theta + \sin^2\theta\right) = 1[/tex]

[tex]r^2 = 1[/tex]

[itex]r[/itex] limits are then from [itex]0[/itex] to [itex]1[/itex]?

The right integral?

[tex]\int_{0}^{2\pi} \int_{0}^{1} r\;ln\left(r^2 + 1\right) dr\;d\theta[/tex]
 
  • #12
Correct, though it might be -1 to 1, spacetiger will give you the final ok.
 
  • #13
The point of all the comments was : DRAW A PICTURE!
[tex]y= \sqrt{1- x^2}[/tex] and [tex]y=-\sqrt{1- x^2}[/tex] both are the same as [tex]x^2= 1- y^2[/tex] or [tex]x^2+ y^2= 1[/tex], the unit circle. Since x ranges from -1 to 1, the region of integration is simply the unit circle. To cover that, [tex]\theta[/tex] ranges from 0 to [tex]2\pi[/tex] and r from 0 to 1.
 

Related to Converting into polar coordinates for integration

1. What are polar coordinates and why are they used for integration?

Polar coordinates are a way of representing points in a plane using a distance from a fixed point (the origin) and an angle from a fixed reference direction. They are often used for integration because they simplify the process of calculating areas and volumes of irregularly shaped objects, such as curves and surfaces.

2. How do I convert cartesian coordinates to polar coordinates?

To convert cartesian coordinates (x, y) to polar coordinates (r, θ), you can use the following formulas: r = √(x² + y²) and θ = tan⁻¹(y/x). Alternatively, you can use trigonometric functions such as sine and cosine to calculate the values of r and θ.

3. What is the purpose of converting into polar coordinates for integration?

The purpose of converting into polar coordinates for integration is to make the integration process simpler and more efficient for certain types of problems. It allows you to avoid complex equations and integrals and instead use basic formulas and trigonometric functions to find the area or volume of a shape.

4. Can polar coordinates be used for integration in 3D space?

Yes, polar coordinates can be used for integration in 3D space. In this case, the coordinates are represented as (r, θ, φ), where r is the distance from the origin, θ is the angle in the xy-plane, and φ is the angle from the positive z-axis. This allows for efficient integration of spherical shapes.

5. Are there any limitations to using polar coordinates for integration?

While polar coordinates can be very useful in many integration problems, they do have some limitations. They are not suitable for all types of shapes, such as rectangular or square objects. Additionally, some integrals may be more complicated to solve in polar coordinates compared to cartesian coordinates.

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