Integral of plane over a sine curve

[Applejacks01]

Hi, I want to verify if my answer to this problem I made up is correct?

Suppose we have a plane z=x+y
Lets find the magnitude of the volume of the space underneath the plane and over the REGION in the xy plane defined by y=sin(x), from 0<=x<=2pi.

So for example, my definition of the "magnitude of the volume" is analogous to saying the magnitude of the integral of sin(x) from 0 to 2pi is 4, not 0 because we took the absolute value of the negative section from pi to 2pi.

So first I integrate x+y dy from 0 to sin(x), then I break the integral
over x up into 2, from 0 to pi, and pi to 2pi. Doing this, I get an answer of 4pi.

Is my solution to what I want correct?

Thanks so much

 

[HallsofIvy]

Hi, I want to verify if my answer to this problem I made up is correct?

Suppose we have a plane z=x+y
Lets find the magnitude of the volume of the space underneath the plane and over the curve in the xy plane y=sin(x), from 0<=x<=2pi.

This makes no sense. There is no volume "over a curve".

so for example, my definition of the "magnitude of the volume" is analogous to saying the magnitude of the integral of sin(x) from 0 to 2pi is 4, not 0 because we took the absolute value of the negative section from pi to 2pi.

So first I integrate x+y dy from 0 to sin(x), then I break the integral
over x up into 2, from 0 to pi, and pi to 2pi. Doing this, I get an answer of 4pi.

Is my solution to what I want correct?

Thanks so much

 

[Applejacks01]

OK all I am saying is integrate using the domain y<=sin(x), 0<=x<=2pi, but solve for the magnitude of the volume.

EDIT; here is the exact region I want to integate over. 0<=y<=sin x from 0<=x<=pi, and sin(x) <=y<= 0 from pi<=x<=2pi