Volume of Container - Triple Integral

[phyzmatix]

Homework Statement

A container has a vertical height of 1m, a circular base with radius 1m and a circular top with radius 2m. Use a triple integral and spherical coordinates to evaluate the volume of the container.

2. The attempt at a solution

If we set up the problem so that the centre of the container's base is at (z,r)=(0,0) it is clear that the region of integration can be viewed as a union of two Type II regions D1 and D2 where

D_1=\{(r,\theta ,z)\|0\leq\theta\leq2 \pi , 0\leq z\leq 1 \mbox{and} 0\leq r\leq 1\}
D_2=\{(r,\theta ,z)\|0\leq\theta\leq2 \pi , 0\leq z\leq 1 \mbox{and} z+1\leq r\leq 2\}

So that

\mbox{Volume of Container}=\int_0^{2\pi}\int_0^1\int_{z+1}^2 r drdzd\theta +\int_0^{2\pi}\int_0^1\int_0^1 r drdzd\theta
=\int_0^{2\pi}\int_0^1 [\frac{1}{2}r^2]_{z+1}^2dzd\theta + \pi
=\frac{1}{2}\int_0^{2\pi}\int_0^1 [4-(z^2+2z+1)]dzd\theta + \pi
=\frac{1}{2}\int_0^{2\pi}\int_0^1 [3-z^2-2z]dzd\theta + \pi
=\frac{1}{2}\int_0^{2\pi} [3z-\frac{1}{3}z^3-z^2]_0^1 d\theta + \pi
=\frac{5}{6}\int_0^{2\pi}d\theta + \pi
=\frac{5}{3}\pi + \pi
=\frac{8}{3}\piThis is the first problem of its kind that I'm attempting and I'm not sure if I've done everything correctly. I'm slightly suspicious of my limits of integration...Could someone please check this and tell me if it's correct or not? I've attached an image of the regions D1 and D2.

Thanks!
phyz

 

[whitay]

You said spherical. So why have you use cylindrical?

 

[phyzmatix]

Thanks

However, as I've gone through the effort of typing this all out, had the question asked for cylindrical coordinates, would this have been correct?

 

[Billy Bob]

D1 is good.

The integration for D2 is "good," but as you suggested, the limits are wrong.

Look at your picture and draw a dot on the z-axis between z=0 and z=1 to represent a generic z value satisfying 0<z<1.

Now draw a horizontal ray emanating from your dot, going to the right. Your ray will enter D2 at r=f(z) and will exit D2 at r=g(z).

Now f(z)=1 (the constant function). What is g(z)? Your drawing is mislabeled, but your work shows that you know it is g(z)=z+1.

Thus, 1<r<z+1 would be the correct limits for r.

 

[phyzmatix]

Oh, ok cool. Thanks for the help!