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

[LCKurtz]

I've been away for a while and missed much of this conversation. But my opinion is that the appropriate way to do this problem is to use a yz plane version of cylindrical coordinates as has been suggested. Something like $$
y = r\cos\theta,~z = r\sin\theta,~ x = x$$

 

[Mark44]

I've been away for a while and missed much of this conversation. But my opinion is that the appropriate way to do this problem is to use a yz plane version of cylindrical coordinates as has been suggested. Something like $$
y = r\cos\theta,~z = r\sin\theta,~ x = x$$

It looks like that's what chetzread did. We both arrived at the same answer.

 

[chetzread]

I've been away for a while and missed much of this conversation. But my opinion is that the appropriate way to do this problem is to use a yz plane version of cylindrical coordinates as has been suggested. Something like $y = r\cos\theta,~z = r\sin\theta,~ x = x$

how to determine whether y is $r\cos\theta$ or y is $r\sin\theta$
or z is $r\cos\theta$ or z is $r\sin\theta$ ...I'm confused ... in this question ... I have zy appear in the integrand . So , i could use (r^2) (cos theta)(sin theta) .I don't know which is $r\cos\theta$ or $r\sin\theta$

 

[LCKurtz]

It doesn't matter which one is the $r\sin\theta$ and which is the $r\cos\theta$.

 

[chetzread]

It doesn't matter which one is the $r\sin\theta$ and which is the $r\cos\theta$.

How if the case where integrand is only z or y ? how to differentiate them ?

 

[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.