Solid bounded by the cylinder (y^2) + (z^2) = 1 , cut by pla

[chetzread]

Homework Statement

By using cylindrical coordinate , evaluate ∫ ∫ ∫ zDv , where G is the solid bounded by the cylinder (y^2) + (z^2) = 1 , cut by plane of y = x , x = 0 and z = 0

I can understand that the solid formed , was cut by x = 0 , thus the base of the solid formed has circle of (y^2) + (z^2) = 1 as base …
The solid formed also cut at z= 0 , does it mean that the sold formed has a base of circle from 0 to π only ?

Homework Equations

The Attempt at a Solution

 

[Mark44]

Homework Statement

By using cylindrical coordinate , evaluate ∫ ∫ ∫ zDv , where G is the solid bounded by the cylinder (y^2) + (z^2) = 1 , cut by plane of y = x , x = 0 and z = 0

I can understand that the solid formed , was cut by x = 0 , thus the base of the solid formed has circle of (y^2) + (z^2) = 1 as base …
The solid formed also cut at z= 0 , does it mean that the sold formed has a base of circle from 0 to π only ?

Homework Equations

The Attempt at a Solution

Your sketch of the region G is not very helpful in seeing what the shape looks like.

does it mean that the sold formed has a base of circle from 0 to π only ?

No. The solid region G is only in the first octant. When you say "from 0 to π only" which angle coordinate do you mean, $\theta$ or $\phi$?

 

[chetzread]

Your sketch of the region G is not very helpful in seeing what the shape looks like.
No. The solid region G is only in the first octant. When you say "from 0 to π only" which angle coordinate do you mean, $\theta$ or $\phi$?

why the solid region G is only in the first octant ? i can't visualise it . Can you explain ?

 

[Mark44]

why the solid region G is only in the first octant ? i can't visualise it . Can you explain ?

The cylinder $y^2 + z^2 = 1$ has its central axis along the x-axis. The bounding planes are x = 0 (the y-z plane), z = 0 (the x-y plane) and the plane y = x. For this last plane draw the line y = x in the x-y plane, and then extend the line up, keeping it parallel to the original line.

Your sketch wasn't detailed enough to do much good (the cylinder should be shown larger), plus you had you axes labeled in an unusual way. That might be throwing you off.

 

[chetzread]

For this last plane draw the line y = x in the x-y plane, and then extend the line up, keeping it parallel to the original line.

from my sketch of diagram , i found that my xy-pane and plane y = x are actually the same plane , so i couldn't imagine the solid formed... what i have in mind now is the y=x plane and the base with circle radius = 1 only ...

 

[LCKurtz]

See if this picture helps you:

 

[chetzread]

See if this picture helps you:
View attachment 108034

thanks for the digram , but , why the y = x plane is limited to first quadrant only ? it can be drawn to up to the plane of positive x and negative y direction , right ?

 

[chetzread]

See if this picture helps you:
View attachment 108034

shouldnt the solid formed look like this ?

 

[Mark44]

shouldnt the solid formed look like this ?

No.
The base of the solid is a triangle. The back side of the solid is the plane x = 0. The front of the solid is the plane y = x. The top is the curved part of the cylinder. The solid itself looks something like half of an orange segment.

 

[chetzread]

No.
The base of the solid is a triangle. The back side of the solid is the plane x = 0. The front of the solid is the plane y = x. The top is the curved part of the cylinder. The solid itself looks something like half of an orange segment.

Why the plane y =x is limited to first octant only ? Can't it be stretched until negative x and y axis?

 

[chetzread]

No.
The base of the solid is a triangle. The back side of the solid is the plane x = 0. The front of the solid is the plane y = x. The top is the curved part of the cylinder. The solid itself looks something like half of an orange segment.

ok , so the limit is
$\int_{0}^{0.5\pi} \int_{0}^1 z \, dx \, rdrd\theta$ ?'
if so , then my ans is -1/8 ...Did my ans match yours?

 

[Mark44]

ok , so the limit is
$\int_{0}^{0.5\pi} \int_{0}^1 z \, dx \, rdrd\theta$ ?'
if so , then my ans is -1/8 ...Did my ans match yours?

The answer shouldn't be negative, and you're supposed to set up an integral usingcylindrical coordinates: r $\theta$, and z.

 

[chetzread]

The answer shouldn't be negative, and you're supposed to set up an integral usingcylindrical coordinates: r $\theta$, and z.

since i project the solid to zy plane , so the third integral should be dx , am i right ?

 

[Mark44]

since i project the solid to zy plane , so the third integral should be dx , am i right ?

No. Neither x nor y should appear in your integral. Also, I don't know why you think you need to project the solid to the y-z plane.

 

[chetzread]

No. Neither x nor y should appear in your integral. Also, I don't know why you think you need to project the solid to the y-z plane.

I project the solid to the y-z plane so that i could use cylindrical coordinate solve the problem , btw when i project it to zy plane , here's what i gt ...
i change z and y to the terms in cylindrical coordinate , the original question is find ∫ ∫ ∫ zDv

 

[Mark44]

I project the solid to the y-z plane so that i could use cylindrical coordinate solve the problem , btw when i project it to zy plane , here's what i gt ...
i change z and y to the terms in cylindrical coordinate , the original question is find ∫ ∫ ∫ zDv

That's not right. In cylindrical coordinates, $\theta$ is an angle in the x-y plane, not in the y-z plane. You need to look at what's going on in the x-y plane.

 

[chetzread]

That's not right. In cylindrical coordinates, $\theta$ is an angle in the x-y plane, not in the y-z plane. You need to look at what's going on in the x-y plane.

Then , it's not possible to use cylindrical coordinate in this question ?

 

[chetzread]

That's not right. In cylindrical coordinates, $\theta$ is an angle in the x-y plane, not in the y-z plane. You need to look at what's going on in the x-y plane.

IMO , the zy plane also can use cylindrical , as long as lines on any 2 axis can form a circle , then we can use cylindrical coordinate

 

[Mark44]

IMO , the zy plane also can use cylindrical , as long as lines on any 2 axis can form a circle , then we can use cylindrical coordinate

I don't know why you would want to do that. Cartesian coordinates are (x, y, z). Cylindrical coordinates are ($r, \theta, z$). If you change the axis that $\theta$ is measured from, you will need to make corresponding changes in the equation of the cylinder and the bounding planes.

 

[chetzread]

I don't know why you would want to do that. Cartesian coordinates are (x, y, z). Cylindrical coordinates are ($r, \theta, z$). If you change the axis that $\theta$ is measured from, you will need to make corresponding changes in the equation of the cylinder and the bounding planes.

ya , i already make that changes in post #11 and my diagram of projection to zy-plane in post #15 , is my concept correct ?

P/s : it's clear that the circle lies on positive/negative y and z axis ... SO , i can use z = rcos theta and y = r sintheta and vice versa , right ?

 

[Mark44]

ya , i already make that changes in post #11 and my diagram of projection to zy-plane in post #15 , is my concept correct ?

No.

Here's what you wrote in post #11:

ok , so the limit is
$\int_{0}^{0.5\pi} \int_{0}^1 z \, dx \, rdrd\theta$ ?'
if so , then my ans is -1/8 ...Did my ans match yours?

I already said this is wrong in post #12. For one thing, your integral has dx (shouldn't be there), dr, and $d\theta$. For another, the integral should be a triple integral, not a double integral as you show.

Finally, no, I haven't worked the integral, but if you're calculating volume, as you are in this problem, you shouldn't get a negative number.

 

[chetzread]

No.

Here's what you wrote in post #11:

I already said this is wrong in post #12. For one thing, your integral has dx (shouldn't be there), dr, and $d\theta$. For another, the integral should be a triple integral, not a double integral as you show.

Finally, no, I haven't worked the integral, but if you're calculating volume, as you are in this problem, you shouldn't get a negative number.

since the circle lies on the zy plane , so i can change the cylindrical coordinate to ( r , θ , x ) , right ? it's just looking at the problem from another point of view , right ? ( for example when the circle lies on the xy -plane , we would use ( r , θ , z )

Sorry , after making correction , my ans $\int_{0}^{0.5\pi} \int_{0}^1 z \, dx \, rdrd\theta$ is 1/ 8

 

[chetzread]

, your integral has dx (shouldn't be there), dr, and dθdθd\theta. For another, the integral should be a triple integral, not a double integral as you show.

dx r dr dtheta is triple integral , right ?

 

[Mark44]

since the circle lies on the zy plane , so i can change the cylindrical coordinate to ( r , θ , x ) , right ? it's just looking at the problem from another point of view , right ?

It's possible to do this, but as I said before, your bounding planes are going to be different. The plane y = x, in Cartesian coordinates, is $\theta = \pi/4$ in cylindrical coordinates. If you change your coordinates to (r, θ, x), then you will have a different equation for the plane y = x.

( for example when the circle lies on the xy -plane , we would use ( r , θ , z )

Sorry , after making correction , my ans $\int_{0}^{0.5\pi} \int_{0}^1 z \, dx \, rdrd\theta$ is 1/ 8

This is not a triple integral.

 

[chetzread]

It's possible to do this, but as I said before, your bounding planes are going to be different. The plane y = x, in Cartesian coordinates, is $\theta = \pi/4$ in cylindrical coordinates. If you change your coordinates to (r, θ, x), then you will have a different equation for the plane y = x.

This is not a triple integral.

sorry , i made the typo again . my final working is $\int_{0}^{0.5\pi} \int_{0}^1 \int_{0}^y z \, dx \, rdrd\theta$ is 1/ 8

As you can see , i use y = x for the limit...I didnt use y = x

 

[Mark44]

sorry , i made the typo again . my final working is $\int_{0}^{0.5\pi} \int_{0}^1 \int_{0}^y z \, dx \, rdrd\theta$ is 1/ 8

As you can see , i use y = x for the limit...I didnt use y = x

Is the answer to this problem in your book?

I don't see how you can make the integral above turn out to 1/8. Please show me your work. In this case, it's OK to take a picture of it, and add it as an attachment, provided that your work is neat enough to read.

I get 1/8 as my answer, but I didn't change the meaning of the cylindrical coordinates.

 

[chetzread]

Is the answer to this problem in your book?

I don't see how you can make the integral above turn out to 1/8. Please show me your work. In this case, it's OK to take a picture of it, and add it as an attachment, provided that your work is neat enough to read.

I get 1/8 as my answer, but I didn't change the meaning of the cylindrical coordinates.

no , i don't have ans... I need someone to verify my answer...

 

[chetzread]

Is the answer to this problem in your book?

I don't see how you can make the integral above turn out to 1/8. Please show me your work. In this case, it's OK to take a picture of it, and add it as an attachment, provided that your work is neat enough to read.

I get 1/8 as my answer, but I didn't change the meaning of the cylindrical coordinates.

Or do you have better way ?

 

[Mark44]

no , i don't have ans... I need someone to verify my answer...

Your work looks OK to me.

Here's the way I did it, without changing all the coordinates around.
The z values range from z = 0 to $z = \sqrt{1 - y^2} = \sqrt{1 - r^2 \sin^2(\theta)}$.
The r values range from r = 0 to the line y = 1 in the x-y plane, so from r = 0 to $r\sin(\theta) = 1$, which is equiv. to $r = \csc(\theta)$
The $\theta$ values range from $\pi/4$ to $\pi/2$.
The integral I start with is:
$$\int_{\theta = \pi/4}^{\pi/2} ~\int_{r = 0}^{\csc(\theta)} ~\int_{z = 0}^{\sqrt{1 - r^2 \sin^2(\theta)}}~zr~dz~dr~d\theta$$
This looks forbidding, but it's actually pretty simple, and ends up with a value of 1/8.

 

[chetzread]

Your work looks OK to me.

Here's the way I did it, without changing all the coordinates around.
The z values range from z = 0 to $z = \sqrt{1 - y^2} = \sqrt{1 - r^2 \sin^2(\theta)}$.
The r values range from r = 0 to the line y = 1 in the x-y plane, so from r = 0 to $r\sin(\theta) = 1$, which is equiv. to $r = \csc(\theta)$
The $\theta$ values range from $\pi/4$ to $\pi/2$.
The integral I start with is:
$$\int_{\theta = \pi/4}^{\pi/2} ~\int_{r = 0}^{\csc(\theta)} ~\int_{z = 0}^{\sqrt{1 - r^2 \sin^2(\theta)}}~zr~dz~dr~d\theta$$
This looks forbidding, but it's actually pretty simple, and ends up with a value of 1/8.

ok , thanks for the working ... Is it wrong with my working above ? Can i do it in the way above ?