Why Does Integrating y Over a Volume Between Two Cylinders Yield Zero?

[semc]

Evaluate \int\int\int y dV where the solid generated lies between the cylinders x²+y²=1 and x²+y²=4, above the xy-plane and below the plane z=x+2.

I wrote out the integral \int\int r*r sin \theta dz dr d\theta and z is integrated from x+2 to 0, r integrated from 2 to 1 and \theta from 2\pi to 0. But I ended up with \frac{-7}{3}(x+2)cos(2\pi - 0) which gives me 0?? I tried visualizing the volume generated and its like a donut with the top cut away at an angle. So how is the volume 0? Did I do something wrong?

 

[Mark44]

Evaluate \int\int\int y dV where the solid generated lies between the cylinders x²+y²=1 and x²+y²=4, above the xy-plane and below the plane z=x+2.

I wrote out the integral \int\int r*r sin \theta dz dr d\theta and z is integrated from x+2 to 0, r integrated from 2 to 1 and \theta from 2\pi to 0. But I ended up with \frac{-7}{3}(x+2)cos(2\pi - 0) which gives me 0?? I tried visualizing the volume generated and its like a donut with the top cut away at an angle. So how is the volume 0? Did I do something wrong?

Your integral doesn't represent volume, so it's possible that its value is 0. If the integrand were 1 instead of 1, then it would represent volume.

 

[semc]

You mean the one I wrote or the equation given?

 

[semc]

*bump*

 

[Pengwuino]

Please do not bump threads.

The integral of JUST dV would represent a volume. However, the integration of a function IN a volume no longer represents the volume.

For example,

\int\limits_0^L {dx}

gives you the length of something from 0-> L, which is just a length of L. However,

\int\limits_0^L {x^2 dx}

no longer gives a length.