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

[LCKurtz]

They are different parameterizations of the same circular region. If you let$$
y = r\cos\theta,~z = r\sin\theta,~ x = x$$or
$$z = r\cos\theta,~y = r\sin\theta,~ x = x$$What is different about them is they aren't the same $\theta$. In the first case $\theta$ measures counterclockwise from the $y$ axis, and in the second case it measures clockwise from the $z$ axis, looking at the zy plane from the positive x direction.

 

[chetzread]

They are different parameterizations of the same circular region. If you let$$
y = r\cos\theta,~z = r\sin\theta,~ x = x$$or
$$z = r\cos\theta,~y = r\sin\theta,~ x = x$$What is different about them is they aren't the same $\theta$. In the first case $\theta$ measures counterclockwise from the $y$ axis, and in the second case it measures clockwise from the $z$ axis, looking at the zy plane from the positive x direction.

so , if the location of angle is not given by the author earlier and the integrand only contain y or z , how to do the question ??

 

[LCKurtz]

so , if the location of angle is not given by the author earlier and the integrand only contain y or z , how to do the question ??

I'm not sure what you are asking. You are given a solid that has a quarter circular projection on the yz plane. If it were on the xy plane you would use polar coordinates if you wanted to integrate over it because it is circular. Same thing for the yz plane. You want polar coordinates. You are free to choose which axis to measure $\theta$ from and which direction is positive. There are many choices of parameterizations that work. The two I mentioned in post #36 are the most obvious "polar coordinate" ones.

 

[chetzread]

See if this picture helps you:
View attachment 108034

why the plane shouldn't look like this (green part)?

 

[LCKurtz]

why the plane shouldn't look like this (green part)?

You can use the projection on the xy plane if you insist. That's what Mark44 showed you in post #29. It's just harder that way, given that you are asked to use cylindrical coordinates. I don't see anything left to do in this thread.

 

[chetzread]

You can use the projection on the xy plane if you insist. That's what Mark44 showed you in post #29. It's just harder that way, given that you are asked to use cylindrical coordinates. I don't see anything left to do in this thread.

No , i am just confused about plane y = x
For plane y = x , it is a triangle that lies on xy plane only(green part) , right ?

Why it's a plane parallel to z as you showed in the diagram ?

 

[LCKurtz]

No , i am just confused about plane y = x
For plane y = x , it is a triangle that lies on xy plane only(green part) , right ?

No. Look at the line y=x in the xy plane. The plane standing vertically on that line in my picture is the plane y=x. And, as you have been told before, planes are not triangles.

Why it's a plane parallel to z as you showed in the diagram ?

Because every point $(x,y,z)$ on that plane satisfies the equation $y=x$ no matter what $z$ is.