Double integral polar/cylindrical coordinates

In summary, the student attempted to find the volume of a solid obtained by rotating a region about a z-axis, but got an incorrect answer due to incorrect use of double integrals.
  • #1
fishturtle1
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Homework Statement


Problem 1: Use double integrals to find the volume of the solid obtained by the rotation of the region:

##\triangle = \left\{ (x, y, z) | x^2 \le z \le 6 - x, 0 \le x \le 2, y = 0 \right\} ## (edit) in the xz-plane about the z axis

Homework Equations


Volume = ##\int_a^b \int_c^d f(r, \theta) r dr d\theta##

The Attempt at a Solution


When I draw this solid, I get a cone who's base is a circle with radius 4.
The height, z, is from z = 4 to z = 6. The sides of the cone are the line 6-x, rotated around the z axis.

##D = \left\{ (r, \theta)_p | r: 0 \rightarrow 4, \theta: 0 \rightarrow 2\pi \right\}##
##f(r, \theta) = 6 - r\cos(\theta)##

V = ##\int_0^{2\pi} \int_0^4 6 - r\cos(\theta) rdrd\theta##
##=\int_0^{2\pi} \int_0^4 6r - r^2\cos(\theta) drd\theta##
##=\int_0^{2\pi} 2r^2 - \frac {r^3}{3}\cos(\theta) d\theta## ...evaluate (r = 4) - (r = 0).
when i typed \right|_0^4 it didn't work..

##= \int_0^{2\pi} 48 - \frac {64}{3}\cos(\theta) d\theta##
##= 48\theta - \frac {64}{3}\sin(\theta)## ...plug in ##\theta = 2\pi## into the expression and subtract ##\theta = 0##.
##= 96\pi + 0 - (0+0)##
##= 96\pi##

but this answer looks way to big for the solid I drew.. where did I first go wrong?
 
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  • #2
fishturtle1 said:

Homework Statement


Problem 1: Use double integrals to find the volume of the solid obtained by the rotation of the region:

##\triangle = \left\{ (x, y, z) | x^2 \le z \le 6 - x, 0 \le x \le 2, y = 0 \right\} ##

Homework Equations


Volume = ##\int_a^b \int_c^d f(r, \theta) r dr d\theta##

The Attempt at a Solution


When I draw this solid, I get a cone who's base is a circle with radius 4.
The height, z, is from z = 4 to z = 6. The sides of the cone are the line 6-x, rotated around the z axis.

##D = \left\{ (r, \theta)_p | r: 0 \rightarrow 4, \theta: 0 \rightarrow 2\pi \right\}##
##f(r, \theta) = 6 - r\cos(\theta)##

V = ##\int_0^{2\pi} \int_0^4 6 - r\cos(\theta) rdrd\theta##
##=\int_0^{2\pi} \int_0^4 6r - r^2\cos(\theta) drd\theta##
##=\int_0^{2\pi} 2r^2 - \frac {r^3}{3}\cos(\theta) d\theta## ...evaluate (r = 4) - (r = 0).
when i typed \right|_0^4 it didn't work..

##= \int_0^{2\pi} 48 - \frac {64}{3}\cos(\theta) d\theta##
##= 48\theta - \frac {64}{3}\sin(\theta)## ...plug in ##\theta = 2\pi## into the expression and subtract ##\theta = 0##.
##= 96\pi + 0 - (0+0)##
##= 96\pi##

but this answer looks way to big for the solid I drew.. where did I first go wrong?

Which is the rotation axis, ##x## or ##z##?
 
  • #3
Ray Vickson said:
Which is the rotation axis, ##x## or ##z##?
The rotation is about the z axis.
 
  • #4
Why did you put ƒ(r,θ)=6-r.cos(θ) as the function? Isn't it just the function of the stright line? And if you put it there it would become a three dimensional equation like y=6-x + 0.z, there you get a plane instead of a cone as you described.
 
  • #5
CollinsArg said:
Why did you put ƒ(r,θ)=6-r.cos(θ) as the function? Isn't it just the function of the stright line? And if you put it there it would become a three dimensional equation like y=6-x + 0.z, there you get a plane instead of a cone as you described.
because I was thinking we had to integrate the line 6-x around the z axis. And to put 6-x in polar form i substituted ##x = \cos\theta##. Are you saying that this means I'm integrating over a plane instead of a cone?
 
  • #6
fishturtle1 said:
When I draw this solid, I get a cone who's base is a circle with radius 4.
The height, z, is from z = 4 to z = 6. The sides of the cone are the line 6-x, rotated around the z axis.
That's not what I get. Why are you ignoring the lower limit ##x^2##?
 
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  • #7
fishturtle1 said:
because I was thinking we had to integrate the line 6-x around the z axis. And to put 6-x in polar form i substituted ##x = \cos\theta##. Are you saying that this means I'm integrating over a plane instead of a cone?
Yes, that's what I meant, remember that the function you plug in there is the function you are integrating (as a "roof") limited by the integral limits you put from each point. I recommend you watching MIT Calculus courses, they are in youtube and are very good ones :)
btw, vela is right, note that you will get a paraboloid cut by a plane, not a cone.
 
  • #8
vela said:
That's not what I get. Why are you ignoring the lower limit ##x^2##?
CollinsArg said:
Yes, that's what I meant, remember that the function you plug in there is the function you are integrating (as a "roof") limited by the integral limits you put from each point. I recommend you watching MIT Calculus courses, they are in youtube and are very good ones :)
btw, vela is right, note that you will get a paraboloid cut by a plane, not a cone.
Thank you for both for the help and correction

I can't remember why I ignored it.. I've redrawn it though and can't think of a reason to ignore it.. Here's my sketch, let me know if its not clear I'll upload again

20171110_191747.jpg


and if this is the case I would integrate the paraboloid and cone separately, but before that, I know you said the surface is a paraboloid, but did you mean a paraboloid in addition to the cone, or just a paraboloid?
 

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  • #9
fishturtle1 said:
Thank you for both for the help and correction

I can't remember why I ignored it.. I've redrawn it though and can't think of a reason to ignore it.. Here's my sketch, let me know if its not clear I'll upload again

View attachment 214789

and if this is the case I would integrate the paraboloid and cone separately, but before that, I know you said the surface is a paraboloid, but did you mean a paraboloid in addition to the cone, or just a paraboloid?

Nice! Honestly, I've never used double integral with polar coordinates to resolve a solid of revolution. But, what I would do, is what you just said, I would first find a way to find the cone and paraboloid equation (because you are not given them) the integrate them separately.

It seems hard, I'm sorry if I'm doing a bad advice couse I'm not sure if it is the way you shoild do it. But notice that when you integrate just the paraboloid you won't get the volume inside it, but outside, maybe you will have to put it downward or substract what you get from a cilinder volume of the same radius when the paraboloid intersect the cone. Then tou would add the volume down the cone.
I'll try to do it myself now, I've just finished Calculus II in the university xD. But I think it would be easer with tripple integrals
 
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  • #10
For every ##r## and ##\theta##, there's a column of height ##f(r,\theta)=z_\text{max}-z_\text{min}##. You want to figure out what that ##f## is for this case and the limits of the integrals.
 
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  • #11
CollinsArg said:
Nice! Honestly, I've never used double integral with polar coordinates to resolve a solid of revolution. But, what I would do, is what you just said, I would first find a way to find the cone and paraboloid equation (because you are not given them) the integrate them separately.

It seems hard, I'm sorry if I'm doing a bad advice couse I'm not sure if it is the way you shoild do it. But notice that when you integrate just the paraboloid you won't get the volume inside it, but outside, maybe you will have to put it downward or substract what you get from a cilinder volume of the same radius when the paraboloid intersect the cone. Then tou would add the volume down the cone.
I'll try to do it myself now, I've just finished Calculus II in the university xD. But I think it would be easer with tripple integrals
I think your advice has been really helpful. Since reading Vela's comment I think I will try to do this in one integral though.

vela said:
For every ##r## and ##\theta##, there's a column of height ##f(r,\theta)=z_\text{max}-z_\text{min}##. You want to figure out what that ##f## is for this case and the limits of the integrals.
$$z_{max} - z_{min} = (6-x) -x^2$$ which in polar coordinates is $$-r^2\cos\theta - r\cos\theta + 6$$

Volume = $$\int_0^{2\pi} \int_0^2 (r^3\cos^2\theta - r^2\cos\theta + 6r) dr d\theta$$
$$= \int_0^{2\pi} [-\frac{r^4}4\cos^2\theta - \frac{r^3}3\cos\theta + 3r^2 ]_0^2 d\theta$$
$$= \int_0^{2\pi} -4\cos^2\theta - \frac83\cos\theta + 12 d\theta$$
$$=[-4(\frac \theta2 + \frac {\sin(2\theta)}{4}) - \frac83\sin\theta + 12\theta]_0^{2\pi}$$
$$=[-\sin(2\theta) - \frac83\sin\theta + 10\theta]_0^{2\pi}$$
$$=20\pi$$

This answer seems reasonable to me.. (I compared it to the volume of the rectangular prism that contains it)... am I right?
 
  • #12
You shouldn't have the cosines in the integral. The expressions ##6-x## and ##x^2## for the upper and lower bounds are valid only in the ##xz##-plane. When you rotate that region around the ##z##-axis, the heights aren't going change with ##\theta##.
 
  • #13
vela said:
For every ##r## and ##\theta##, there's a column of height ##f(r,\theta)=z_\text{max}-z_\text{min}##. You want to figure out what that ##f## is for this case and the limits of the integrals.

I don't understand why the function would be a difference between the two z's, is there a visual way to see this?
 
  • #14
vela said:
You shouldn't have the cosines in the integral. The expressions ##6-x## and ##x^2## for the upper and lower bounds are valid only in the ##xz##-plane. When you rotate that region around the ##z##-axis, the heights aren't going change with ##\theta##.
So looking at the previous example, they used cylindrical coordinates..

So
##f(r,\theta) = z_{max} - z_{min}##
##z_{max} = 6-x## But x = r since y = 0.
So ##z_{max} = 6 - r##.

##z_{min} = x^2 = r^2## since x = r.

So ##f(r,\theta) = 6 - r - r^2##.

So Volume = ##\int_0^{2\pi} \int_0^2 (6-r-r^2) r dr d\theta##
 
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  • #15
fishturtle1 said:
So looking at the previous example, they used cylindrical coordinates..

So
##f(r,\theta) = z_{max} - z_{min}##
##z_{max} = 6-x## But x = r since y = 0.
So ##z_{max} = 6 - r##.

##z_{min} = x^2 = r^2## since x = r.

So ##f(r,\theta) = 6 - r - r^2##.

So Volume = ##\int_0^{2\pi} \int_0^2 (6-r-r^2) r dr d\theta##

I got 8π/3 with that integral, but doing it the way I told you before I got 32π/3. I'm not sure if it's right, anyways I still don't understand, why the function is the difference of z's, though I understood the procedure you did afterwards, but I can't connect why should the function be that difference.
 
  • #16
CollinsArg said:
I got 8π/3 with that integral, but doing it the way I told you before I got 32π/3. I'm not sure if it's right, anyways I still don't understand, why the function is the difference of z's, though I understood the procedure you did afterwards, but I can't connect why should the function be that difference.
I just did it and got ##\frac {32\pi}{3}## for the integral in post #14

Volume = ##\int_0^{2\pi} \int_0^2 (6r - r^2 - r^3) dr d\theta##
##= \int_0^{2\pi} [3r^2 - \frac {r^3}{3} - \frac {r^4}{4}]_0^2 d\theta##
##=\int_0^{2\pi} 12 - \frac 83 - 4 d\theta##
##= \frac{16}{3} \theta ]_0^{2\pi}##
##= \frac {32\pi}{3}##
 
  • #17
fishturtle1 said:
I just did it and got ##\frac {32\pi}{3}## for the integral in post #14

Volume = ##\int_0^{2\pi} \int_0^2 (6r - r^2 - r^3) dr d\theta##
##= \int_0^{2\pi} [3r^2 - \frac {r^3}{3} - \frac {r^4}{4}]_0^2 d\theta##
##=\int_0^{2\pi} 12 - \frac 83 - 4 d\theta##
##= \frac{16}{3} \theta ]_0^{2\pi}##
##= \frac {32\pi}{3}##
Oh yes, I just forgot a four,because I had taken r2 as common factor. Then it seems it's right :)
 
  • #18
CollinsArg said:
I don't understand why the function would be a difference between the two z's, is there a visual way to see this?
Plot ##z=6-x## and ##z=x^2## in the ##xz##-plane. Remember that you're rotating the bounded region around the ##z##-axis.
 

Related to Double integral polar/cylindrical coordinates

1. What is a double integral in polar/cylindrical coordinates?

A double integral in polar/cylindrical coordinates is a mathematical concept used to calculate the volume or area of a region in two-dimensional or three-dimensional space. It involves integrating a function over a specific region, where the coordinates are defined in terms of polar or cylindrical coordinates instead of the traditional Cartesian coordinates.

2. How is the double integral expressed in polar/cylindrical coordinates?

In polar/cylindrical coordinates, the double integral is expressed as ∫∫f(r,θ)r dr dθ, where r represents the distance from the origin and θ represents the angle of rotation.

3. What is the difference between polar and cylindrical coordinates?

While both polar and cylindrical coordinates use a radius and angle to define a point in space, polar coordinates are used in two-dimensional space while cylindrical coordinates are used in three-dimensional space. In cylindrical coordinates, an additional coordinate, z, is added to represent the height or depth of a point.

4. What types of problems can be solved using double integrals in polar/cylindrical coordinates?

Double integrals in polar/cylindrical coordinates are useful for solving problems involving areas, volumes, and physical systems with circular or cylindrical symmetry. They are commonly used in physics, engineering, and other fields to calculate the mass, charge, or other physical quantities in a given region of space.

5. How do I convert a double integral from Cartesian coordinates to polar/cylindrical coordinates?

To convert a double integral from Cartesian coordinates to polar/cylindrical coordinates, we use the following substitution: x = rcos(θ) and y = rsin(θ). This allows us to express the function in terms of r and θ instead of x and y, and then we can use the polar/cylindrical double integral formula to solve the problem.

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