Bassman's question at Yahoo Answers regarding a solid of revolution

[MarkFL]

Here is the question:

Calculus Volume of rotation question?

Hi, I am having difficulty with this volume of rotation problem

y=2-1/2x, y=0, x=1, x=2

The answer is 19pi/12 in the book but I when I try to solve the problem I cannot
work it out correctly. How would one go about working out this problem?
Thanks for you're helpAdditional Details:

I forgot to add that it is rotating about x-axis

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello bassman,

The volume we are being asked to compute is that of the frustum of a cone, and the formula for such a solid is derived using the calculus here:

http://mathhelpboards.com/math-notes-49/volumes-pyramids-6131.html

$$V=\frac{\pi h}{3}\left(R^2+Rr+r^2 \right)$$

In this problem, we have:

$$h=1,\,R=\frac{3}{2},\,r=1$$

Hence:

$$V=\frac{\pi}{3}\left(\frac{9}{4}+\frac{3}{2}+1 \right)=\frac{\pi}{3}\left(\frac{19}{4} \right)=\frac{19\pi}{12}$$

Thus, we know the answer given by the book is correct. So, let's work this in the manner you are expected to use.

Let's use the disk method. The volume of an arbitrary disk is:

$$dV=\pi r^2\,dx$$

where:

$$r=y=2-\frac{1}{2}x=\frac{4-x}{2}$$

and so we have:

$$dV=\pi\left(\frac{4-x}{2} \right)^2\,dx=\frac{\pi}{4}(4-x)^2\,dx$$

Adding the disks by integrating, we may state:

$$V=\frac{\pi}{4}\int_1^2(4-x)^2\,dx$$

Using the substitution:

$$u=4-x\,\therefore\,du=-dx$$

we may state:

$$V=\frac{\pi}{4}\int_2^3 u^2\,du=\frac{\pi}{12}\left[u^3 \right]_2^3=\frac{\pi}{12}\left(3^3-2^3 \right)=\frac{19\pi}{12}$$