# Polar Coordinate Integration

• August 14th 2006, 05:36 PM
jcarlos
Polar Coordinate Integration
Hello,

Having problems setting integration limits due to symmetry.

I have two questions:

One asks to find the volume of the solid formed by the interior of the circle
r = cos(theta) capped by the plane z = x.

Because x = rcos(theta) and r = cos(theta), we have z = (cos(theta))^2
= (1+ cos2(theta))/2, which is the function we integrate.

The volume we wish to find is the right hand side of the lemniscate which is enclosed inside the circle.

For limits for the dr expression we can integrate from b = 1 to a = 0.
For the d(theta) limits I get confused. I was told to integrate the smallest interval possible and multiply by any symmetry factors. With that said, I can take beta = pi/2 and alpha = 0 and multiply by two due to symmetry about the x-axis. However, I know this is not the smallest interval as there must be a ray that caps the lemniscate function. I know how to find this ray, essentially a value for theta, by setting functios r equal to each other and finding that theta. However, I have an expression in terms of r and in terms of z.

My confusion in this next question is much the same as the above:
The question asks to find the volume of the solid based on the interior of the cardioid r = 1 + cos(theta), capped by the cone z = 2 - r.

We have z = 2 - (1 + cos(theta)) = 1 - cos(theta). Essentially another cardioid with the same size but with the cusp pointing in the opposite direction. The graph I get is basically an infinity sign along the y-axis, that is with rays pi/2 and 3pi/2, and with vertices (1,0) and (-1,0).

Again I get confused by symmetry due to the negative and positive contributions of the polar graph. For the d(theta) limits I believe I can integrate from alpha = 0 to beta = pi/2 and multiply by four due to symmetry. I get confused for the dr limits as the solid formed by the two cardioids expands over all four quadrants. Do I integrate from a = o to b = 2 and multiply by two due to symmetry? Two is the intersection point of the
1 + cos (theta) cardiod on the x-axis. I really get confused with this question as I imagine you can integrate from 0 to pi/2, pi/2 to pi, pi to 3pi/2 and 3pi/2 to 2pi breaking up the four quadrants and adding up the volumes of the four regions of the solid.

I would love any hints and any generalizations about symmetry considerations with polar coordinates.

Thank you.
• August 14th 2006, 06:20 PM
ThePerfectHacker
Quote:

Originally Posted by jcarlos
Hello,

Having problems setting integration limits due to symmetry.

I have two questions:

One asks to find the volume of the solid formed by the interior of the circle
r = cos(theta) capped by the plane z = x.

The region of integration is demonstrated below.

To find the surface area you need to find,
$\int_D \int \sqrt{1+f_x^2+f_y^2} dA$
Given the surface,
$z=x$ we find that,
$f_x=1$ and $f_y=0$,
Thus,
$\sqrt{1+f_x^2+f_y^2}=\sqrt{1+1^2+0^2}=\sqrt{2}$
Thus,
$\int_D \int \sqrt{2} dA=\sqrt{2}\int_D\int dA$.
But,
$\int_D \int dA$ is the area of the region. Which is,
$\frac{\pi}{4}$
Thus,
$\frac{\pi\sqrt{2}}{4}$
• August 14th 2006, 06:43 PM
ThePerfectHacker
Quote:

Originally Posted by jcarlos

My confusion in this next question is much the same as the above:
The question asks to find the volume of the solid based on the interior of the cardioid r = 1 + cos(theta), capped by the cone z = 2 - r.

We have z = 2 - (1 + cos(theta)) = 1 - cos(theta). Essentially another cardioid with the same size but with the cusp pointing in the opposite direction. The graph I get is basically an infinity sign along the y-axis, that is with rays pi/2 and 3pi/2, and with vertices (1,0) and (-1,0).

The region of integration is demonstrated below.
To simplify the problem divide it into to parts. Calculate the right region and then add the left region.
The cone is, $z=2-(x^2+y^2)$??? But that is a parabolid :eek: So I presume you meant to say $z=2-\sqrt{x^2+y^2}$. Thus, you need to find,
$\int_{A_1} \int 2-x^2-y^2 dA +\int_{A_2} \int 2-x^2-y^2 dA$ after the substitution $x^2+y^2=r^2$ you end with (and remember to multiply by $r$ again),
$\int_{A_1} \int (2-r)r dr d\theta +\int_{A_2}\int (2-r)r dr d\theta$
Now you need to set the limits. Which are,
$\int_{3\pi/2}^{\pi/2} \int_0^{1-\cos \theta} (2-r)r dr d\theta +\int_{\pi/2}^{3\pi/2} \int_0^{1+\cos \theta} (2-r)r dr d\theta$
• August 14th 2006, 08:24 PM
jcarlos
Polar coordinate Integration
Perfect Hacker,

Thank you for your explanations the question with the cardioids is now clear.
However, I do not understand your reply to my first question. I don't see what the surface area has to do with the solid region inside the circle
r = cos(theta) and the plane z = x.
The polar coordinate integration requires dA = rdrd(theta), which is not used in your explanation.

The graph I obtain has the right hand of the lemniscate inside of the circle.

I used z = x = rcos(theta)
Therefore, z = cos(theta)cos(theta) = (1 + cos2(theta))/2. This is the function we integrate right.
So the limits can be from r = cos(theta) to r = 0 right?!
I am not sure about the other limits though, would it be from pi/2 to 0 and multiply by two because of symmetry?

Thanks again.
• August 14th 2006, 11:33 PM
jcarlos
Polar Coordinate Integration
Perfect Hacker,

I finished doing the question with the two cardioids. Unless I have made some mistakes, working out the integrals for 2r - r^2 was very labourious. I got an answer of -28/9 for the volume of the solid formed by the two cardioiods. Does a negative volume make any sense?

Thank you again.
• August 15th 2006, 07:58 AM
ThePerfectHacker
Forgive me. In the first question I assumed you were speaking of surface area not volume. I respond back later.
• August 15th 2006, 08:05 AM
ThePerfectHacker
Quote:

Originally Posted by jcarlos
Perfect Hacker,

I finished doing the question with the two cardioids. Unless I have made some mistakes, working out the integrals for 2r - r^2 was very labourious. I got an answer of -28/9 for the volume of the solid formed by the two cardioiods. Does a negative volume make any sense?

Thank you again.

No, it cannot be negative unless the surface goes below the xy-plane. It you visualize the surface $z=2-\sqrt{x^2+y^2}$ is it above. It is very possible you make a mistake in this problem because it is very long computation. It is also possible that I made a mistake with the first integral by writing the angles bound incorrecty, I shall check that. I belive it might be simpler to express this as 4 integral. Divide that region into 4 parts.
---
I believe the integral should have been,
$
\int_{-\pi/2}^{\pi/2} \int_0^{1-\cos \theta} (2-r)r dr d\theta +\int_{\pi/2}^{3\pi/2} \int_0^{1+\cos \theta} (2-r)r dr d\theta
$

When we, use the first iteration we get,
$\int_{-\pi/2}^{\pi/2} (1-\cos \theta)^2-\frac{1}{3}(1-\cos \theta)^3 d\theta \approx .5388$ (I used software for this part).

On the second integral,
$\int_{\pi/2}^{3\pi/2}(1+\cos \theta)^2-\frac{1}{3}(1+\cos \theta)^3 d\theta\approx .5388$
---
I believe when you mentioned symettry was a good idea.
The cone, $f(x,y)=2-\sqrt{x^2+y^2}$ is symettric because $f(-x,-y)=f(x,y)$.
Thus, you good have calculated the right part and then multipled by two, namely,
$2\int_{\pi/2}^{3\pi/2}(1+\cos \theta)^2-\frac{1}{3}(1+\cos \theta)^3 d\theta$
• August 15th 2006, 10:27 PM
jcarlos
Polar coordinate integration
Perfect Hacker,

Thank you again for the question with the cardiods. The limits you proposed the first time were incorrect. With the new limits you suggested, I got the same volume as you did by either adding up the left and the right side, or by computing either and multiplying by two due to symmetry.
Like I said before, quite a labourious question with all the trigonometric substitutions.

I am confident that I did the other question correctly, the one you misunderstood as surface area instead of volume.

For d(theta) limits I integrated from 0 to pi/2 and multiplied by two due to symmetry, and for dr limits, I integrated from 0 to cos(theta). The function I integrated was (1+ cos2(theta))/2. Please correct me if I am wrong.

Thanks again.
• August 16th 2006, 07:51 AM
ThePerfectHacker
Quote:

Originally Posted by jcarlos
For d(theta) limits I integrated from 0 to pi/2 and multiplied by two due to symmetry, and for dr limits, I integrated from 0 to cos(theta). The function I integrated was (1+ cos2(theta))/2. Please correct me if I am wrong.

Seems right. Just let me tell you a story about symettry. You should not rely on it so much. Because it is based on the symettry of the region (like here) AND the symettry of the solid above the region. Once you have that then you can proceede with symettry.