Question

My question is number 14-25. My professor gave me the answers, but wants to see my work. I have no idea where to even start, can anyone help me?
http://www.stewartcalculus.com/data/ESSENTIAL%20CALCULUS%202e/upfiles/instructor/eclt_ax_0702.pdf

Answer 1

Which one from that section?

Answer 2

16 please

Answer 3

So we are rotating \(f(x)=\frac{x}{4}\) where \(0\leq x\leq8\) around the line \(x=8\) Right?

Answer 4

Sure, I honestly have no idea.

Answer 5

We can shift \(f(x)\) 8 units to the left to create the transformed function, \(g(x)\), and so then we have a rotation of \(g(x)\) about the y-axis. Let us find out \(g(x)\).
\[g(x)=f(x+8)=\frac{x+8}{4}=\frac{x}{4}+2\]
So then we have reduced the question to revolving the function \(g(x)\) where \(-8\leq x\leq0\) around the y axis. Let us convert \(y=g(x)\) into terms of x:
So we have:
\[\eqalign{
&y=\frac{x}{4}+2 \\
&y-2=\frac{x}{4} \\
&4(y-2)=x \\
&x=4y-8 \\
&h(y)=4y-8 \\
}\]
So we can also redefine the limits of integration from \((-8\leq x\leq0)\) to \((0\leq y\leq2)\)
So far, we have changed the question to one of rotating \(h(y)\) around the y-axis from \((0\leq y\leq2)\)

Cool so far?

Answer 6

Then, we have the equation that rotating \(h(y)\) around the y-axis from \(a\leq y\leq b\)
Now for some fun stuff ;)

This is the same as setting up the integral of:
\[\pi\int^b_a{[h(y)]^2dy}\]

Then we can develop the equation (using \(V\) for volume):
\[V=\pi\int^2_0[4y-8]^2dy=\pi\int^2_0{[4(y-2)]^2dy=\pi\int^2_016(y-2)^2dy}\]
We can evaluate this definite integral like so:
\[V=16\pi\int^2_0{(y-2)^2dy}=16\pi\int^2_0{u^2dy}\]
\[\eqalign{
&u=y-2 \\
&du=(1) \\
&du=dx \\
}\]
\[V=16\pi\int^2_0{u^2du}=16\pi\left.\left(\frac{u^3}{3}\right)\right|^2_0=16\pi\left.\left[\frac{(y-2)^3}{3}\right]\right|^2_0=16\pi(\mu)\]
Ill compute \(\mu\) separately just to make it neater:
\[\mu=\left.\left[\frac{(y-2)^3}{3}\right]\right|^2_0=\left[\frac{(2-2)^3}{3}-\frac{(0-2)^3}{3}\right]=\left[0-\frac{(-2)^3}{3}\right]=\left(-\frac{-8}{3}\right)=\frac{8}{3}\]
SO we can return to the original question:
\[V=16\pi\mu=16\pi\left(\frac{8}{3}\right)=\frac{128\pi}{3}\]

Answer 7

PHEW! And that's your answer, I believe. Is it the same answer?

Answer 8

That is the correct answer, thank you so much!!!!

Answer 9

Anytime :)