Is this double integral set up correctly?

[G01]

1.Set up the integral to Find the volume enclosed by the cylinder x^2 +y^2 = 1 x=0 and z=y


3. The area to integrate over is the part of x^2 + y^2 =1 above the x axis. X goes from -1 to 1 and y goes from 0 to sqrt(1-x^2) So the integral should be:

$$\int^1_{-1} \int^{\sqrt(1-x^2)}_0 y dy dx$$

 

[Swapnil]

The way you phrased the question, I don't think that the volume is above the x-axis (or the x-plane). The x=0 plane is the yz plane, so it is actually cutting your cylinder in-half vertically and the the z=y plane is slashing your cylinder diagonally.

 

[HallsofIvy]

1.Set up the integral to Find the volume enclosed by the cylinder x^2 +y^2 = 1 x=0 and z=y


3. The area to integrate over is the part of x^2 + y^2 =1 above the x axis. X goes from -1 to 1 and y goes from 0 to sqrt(1-x^2) So the integral should be:

$$\int^1_{-1} \int^{\sqrt(1-x^2)}_0 y dy dx$$

Please go back and recheck the statement of the problem. x²+ y^{2[/suo] is an infinite cylinder and x= 0 is a plane crossing the x-axis. Together they bound an infinite "half-cylinder". z= y is a plane that crossing that half-cylinder. The three together do NOT bound a region of finite volume. The region either above or below z= y does not have finite volume. If the second equation were z= 0, then it would make sense.}