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

1. Oct 26, 2016

1. The problem statement, all variables and given/known data
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 ?

2. Relevant equations

3. The attempt at a solution

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2. Oct 26, 2016

### Staff: Mentor

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 $\phi$?

3. Oct 26, 2016

why the solid region G is only in the first octant ? i cant visualise it . Can you explain ?

4. Oct 26, 2016

### Staff: Mentor

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.

5. Oct 26, 2016

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 .....

6. Oct 26, 2016

### LCKurtz

See if this picture helps you:

7. Oct 26, 2016

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 postive x and negative y direction , right ?

8. Oct 26, 2016

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9. Oct 26, 2016

### Staff: Mentor

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.

10. Oct 27, 2016

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

11. Oct 27, 2016

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?

12. Oct 27, 2016

### Staff: Mentor

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

13. Oct 27, 2016

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

14. Oct 27, 2016

### Staff: Mentor

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.

15. Oct 27, 2016

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

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16. Oct 27, 2016

### Staff: Mentor

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.

17. Oct 27, 2016

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

18. Oct 27, 2016

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

19. Oct 27, 2016

### Staff: Mentor

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.

20. Oct 27, 2016

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 ???

21. Oct 27, 2016

### Staff: Mentor

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.

22. Oct 27, 2016

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

Last edited: Oct 27, 2016
23. Oct 27, 2016

dx r dr dtheta is triple integral , right ?

24. Oct 27, 2016

### Staff: Mentor

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.

25. Oct 27, 2016

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