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 However, the integration of a function IN a volume no longer represents the volume.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.