Double Integral with Polar Coordinates

In summary, \int_{0}^{cos\vartheta} r dr d \varthetaIs the region for the inner integral from 0 to x, or from x = 0 to x = cos\vartheta?The inner integral covers the region from 0 to cos\vartheta.
  • #1
cse63146
452
0

Homework Statement



[tex]\int^{0}_{-3}\int^{\sqrt{9 - x^2}}_{- \sqrt{9 - x^2}} \sqrt{1 + x^2 + y^2} dy dx[/tex]

Homework Equations



x = rcos(theta)
y = rsin(theta)


The Attempt at a Solution



By making [tex]\sqrt{9 - x^2} = y[/tex] then changing it to polar coordinates, I got r to be +/-3

but I'm not sure how to find what theta is bounded by. From what I read, since its [tex]\sqrt{1 + x^2 + y^2}[/tex] it's in the first quadrant. Not sure what to do now.
 
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  • #2
Your region of integration is the left half of a circle of radius 3, centered at the origin. I would let r run from 0 to 3, theta from pi/2 to 3pi/2.
 
  • #3
How did you know that theta runs from pi/2 to 3pi/2?
 
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  • #4
cse63146 said:
How did you know that theta runs from pi/2 to 3pi/2?
He just told you:
Mark44 said:
Your region of integration is the left half of a circle of radius 3, centered at the origin.
[itex]\theta= 0[/itex] is in the direction of the positive x-axis, [itex]\theta= \pi/2[/itex] points in the direction of the positive y-axis, [itex]\theta= \pi[/itex] points in the direction of the negative x-axis, and [itex]\theta= 3\pi/2[/itex] points in the direction of the negative x-axis. The cover the left half of the circle, you go counterclockwise from the positive y-axis to the negative x-axis: from [itex]\pi/2[/itex] to [itex]3\pi/2[/itex].
 
  • #5
HallsofIvy said:
The cover the left half of the circle, you go counterclockwise from the positive y-axis to the negative x-axis: from [itex]\pi/2[/itex] to [itex]3\pi/2[/itex].

Can't believe I didn't realize that; I'm such an idiot. Thank you both.

I'm probably wrong, but wouldn't [itex]\theta= 3\pi/2[/itex] point to the negative y - axis, since it's the left half of a circle?
 
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  • #7
So I tried another one, but the book says I'm wrong:


[tex]\int^{1}_{0}\int^{x}_{0} \frac{x + y}{\sqrt{x^2 + y^2}} dx dy = \int^{\pi / 2}_{0}\int^{1}_{0} \frac{rcos\vartheta + rsin \vartheta}{\sqrt{r^2 cos^2\vartheta + r^2 sin^2 \vartheta}} r dr d \vartheta = \int^{\pi / 2}_{0}\int^{1}_{0} \frac{rcos\vartheta + rsin \vartheta}{r} r dr d \vartheta = 1/2 \int^{\pi / 2}_{0} (sin\vartheta - cos \vartheta ) r^2 |^1_0 d \vartheta = [/tex]

[tex] 1/2 \int^{\pi / 2}_{0} sin \vartheta - cos \vartheta d\vartheta[/tex]

when I change the boundry for x to polar coordinates, would it go from 0 - 1 or 0 - rcos?
 
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  • #8
Your limits on the the first two iterated integrals are wrong. One of the key points in switching from Cartesian to polar is understanding exactly what the region is over which integration takes place.

For the first integral, it looks like you copied the order of integration incorrectly, and it should by dy, then dx, not the other way around as you have them.

Can you describe the region on which integration takes place in the first integral? It's a very simple geometric object. When you have that figured out, you should be able to get the right limits on your polar integral.
 
  • #9
Mark44 said:
For the first integral, it looks like you copied the order of integration incorrectly, and it should by dy, then dx, not the other way around as you have them.

It's the way in my book, unless I have to change them.

For the first integral, y is bounded between 0 - 1, while x is bounded on the positive x-axis (since it's 0 - x).

dx dy = r dr d(theta) or is it dy dx = r dr d (theta)?
 
  • #10
cse63146 said:
It's the way in my book, unless I have to change them.

For the first integral, y is bounded between 0 - 1, while x is bounded on the positive x-axis (since it's 0 - x).
That's a typo in your book. If it were as you are interpreting it, the region would be infinitely large. x is bounded by what, x?
That's the same as saying that the region for the inner integral is from x = 0 to x = x, and for the outer integral, y = 0 to y = 1.

I'm sure this is what they meant:
[tex]\int_{x = 0}^1 \int_{y = 0}^x <your f(x, y)> dy dx[/tex]
and got the differentials in the wrong order.
cse63146 said:
dx dy = r dr d(theta) or is it dy dx = r dr d (theta)?
Yes.
As far as areas are concerned dx*dy = dy*dx, and they both equal r*dr*d(theta). The difference is that the order of dx and dy controls the order in which integration is done.
 
  • #11
so y is bounded between 0 and x, while x is bounded by 0 and 1. That does make more sense. It looks like it's in the first quadrant. So changing it to polar coodrinates:

[tex]\int_{0}^{\pi / 4} \int_{0}^{cos\vartheta} r dr d \vartheta[/tex]
 
  • #12
The limits look good, so I trust that you understand the integration region. Don't forget your integrand, though. The integral you have just computes the area of the region, and that's not what you want.
 
  • #13
Thanks for your help. Just got one more question. I'm wondering if how I changed the boundries from rectangular coordiantes to polar coordinates is the correct way, and I didnt get the right answer by some fluke.

in [tex]\int_{0}^{\pi / 4} \int_{0}^{cos\vartheta} r dr d \vartheta[/tex]

I did this: y = x => rsin = rcos => sin = cos (and this can only be true when the angle is (pi/4)

for [tex]\int^{0}_{-3}\int^{\sqrt{9 - x^2}}_{- \sqrt{9 - x^2}} \sqrt{1 + x^2 + y^2} dy dx[/tex]

I let [tex]y = \sqrt{9 - x^2}\Rightarrow rsin = \sqrt{9 - r^2 cos^2}[/tex] which eventually gave me r = +/- 3.

Sorry for being so annoying.
 
  • #14
On thinking about it some more, your upper limit of integration needs to be r = 1/cos(theta). The limits of integration x = 0 to x = 1 and y = 0 to y = x represent a right triangle with vertices at (0,0), (1, 0) and (1, 1). It's isosceles with angles of 45 degrees, for which sin(theta) = cos(theta) or tan(theta) = 1. In the polar integration r ranges from 0 to the line x = 1, or rcos(theta) = 1, or r = 1/cos(theta).

For the other (actually the first integral in this thread), the region is the left half of a circle whose radius is 3 and where theta ranges between pi/2 and 3pi/2 (already discussed).

In converting from one type of iterated integral to another, the limits of integration depend completely on the region, so if you don't understand what the region looks like, you won't get them right in the new integral.
 
  • #15
but I thought r ranges from 0 to the line y=x (or is the same as y = x = 1?)
 
  • #16
If r varied from 0 to 1 and theta varied from 0 to pi/4, you would have a region that is the sector of a circle. It would have the curvature because r would include all values between 0 and 1 on this eighth of a circle.

Your region, on the other hand, is a right triangle. Do you see the difference?
 
  • #17
cse63146 said:
but I thought r ranges from 0 to the line y=x (or is the same as y = x = 1?)
Not the same. y = x = 1 is the point (1, 1).
 

1. How do you convert a double integral to polar coordinates?

To convert a double integral to polar coordinates, you need to use the following formula:
∫∫ f(x,y) dA = ∫∫ f(r cos θ, r sin θ) r dr dθ
Where r represents the distance from the origin and θ represents the angle in polar coordinates. You can use this formula to change the limits of integration and the differential element in order to solve the integral in polar coordinates.

2. What is the purpose of using polar coordinates in a double integral?

Polar coordinates are useful in solving double integrals when the region of integration has circular or symmetric boundaries. It allows us to express the integrand in simpler terms and makes the integration process more manageable. It also helps in visualizing the geometric interpretation of the integral in terms of radius and angle.

3. Can you use polar coordinates for any type of double integral?

No, you cannot use polar coordinates for any type of double integral. Polar coordinates are only useful for solving integrals over regions with circular or symmetric boundaries. If the region of integration has rectangular or irregular boundaries, it is better to use rectangular coordinates.

4. How do you set up the bounds for a double integral in polar coordinates?

To set up the bounds for a double integral in polar coordinates, you need to consider the shape and symmetry of the region of integration. If the region is circular, the bounds for r will be from 0 to the radius of the circle, and the bounds for θ will be from 0 to 2π. If the region is symmetric about the x-axis, the bounds for θ will be from 0 to π, and for r, it will depend on the curve that forms the upper boundary of the region.

5. Can you evaluate a double integral in polar coordinates without converting it?

Yes, you can evaluate a double integral in polar coordinates without converting it to rectangular coordinates. However, it may be more challenging to visualize the region of integration and set up the bounds in polar coordinates. It is recommended to convert the integral to rectangular coordinates if the region of integration has non-circular or non-symmetric boundaries.

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