Volume by Cross Sections

[Guest]

hi... i was sick during my lecture on cross sections, so i was wondering if u guys could tell me if i did the first 2 right? Also, my third answer came out negative, so i guess something went wrong with it?

y=x^2, y=0, x=1 about x-axis

so i just did pi * S (X^2)^2 dx= pi/5 (from o to 1)

y=1/x y=0 x=0.1 x=1 about x-axis

pi* S (1/x)^2dx= 9 pi (from 0.1 to 1)

This is the one i know i goofed for sure.

Find the volume when y=x^2, y=4x is revolved about x=5
so i rewrote the equations and got x=y^.5 and x=y/4 Then I drew a picture and found the big radius to be 5-y/4 and the little radius to be 5-y^.5 ?? then i squared both and subtracted and got a really big negative number. What did i do incorrectly? Thanks!
(sorry i dont know how to use the integral symbol so its probably hard to follow!) :?

 

[galactus]

#3.

You can use shells or washers. Take your pick.

\(\displaystyle \L\\y=x^{2}\,\ y=4x\,\ about\,\ x=5\)

When you use shells, the crossections are parallel to the axis about which it's being revolved. Since we're revolving about x=5, that's vertical, so the cross sections will be stacked up along the x-axis.

To find the limits of integration, equate the functions.

\(\displaystyle \L\\x^{2}=4x\,\ x=0,4\)

\(\displaystyle \L\\2{\pi}\int_{0}^{4}(x-5)(x^{2}-4x)dx\)

Now, try to set it up using slices. If you set it up correctly, you'll get the same answer.

 

[soroban]

Hello, aswimmer113!

\(\displaystyle y \,=\,x^2,\;y\,=\,0,\;x\,=\,1\,\text{ about }x\text{-axis}\)

so i just did: \(\displaystyle \L\,\pi\int^{\;\;\;1}_0 (x^2)^2\,dx \;= \;\frac{\pi}{5}\)

Correct . . . good work!

\(\displaystyle y\,=\,\frac{1}{x},\;\;y\,=\,0,\;\;x\,=\,0.1,\;\; x=1\,\text{ about }x\text{-axis}\)

\(\displaystyle \L V \;= \;\pi\int^{\;\;\;1}_{0.1}\left(\frac{1}{x}\right)^2\, dx\;= \;9\pi\)

Also correct . . . nice going!

\(\displaystyle y\,=\,x^2,\;\;y\,=\,4x\,\text{ about }x\,=\,5\)

Your method should have worked.
The intersection is \(\displaystyle (4,16)\) . . . Did you use \(\displaystyle y\)-iimits?

Your set-up should have looked like this:
\(\displaystyle \L\;\;\;V \;= \;\int^{\;\;\;16}_0\left[\left(5\,-\,\frac{1}{4}y\right)^2\,-\,\left(5\,-\,y^{\frac{1}{2}}\right)^2\right]\,dy\)

\(\displaystyle \;\;\)which takes a lot of work . . . and comes out to \(\displaystyle \L 64\pi\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I did it by cylindrical shells: \(\displaystyle \L\,V \;=\;2\pi\int^{\;\;\;b}_a(radius)(height)\,dx\)

So we have: \(\displaystyle \L\,V\;=\;2\pi\int^{\;\;\;4}_0\(5\,-\,x)(4x\,-\,x^2)\,dx\)
\(\displaystyle \L\;\;=\;2\pi\int^{\;\;\;4}_0\left(20x\,-\,9x^2\,+\,x^3)\,dx\)

\(\displaystyle \;\;=\; 2\pi\left(10x^2\,-\,3x^3\,+\,\frac{1}{4}x^4\right) \,\L|\)\(\displaystyle \begin{array}{ccc}4\\:\\0\end{array}\)

\(\displaystyle \;\;=\;2\pi\left[(10\cdot16\,-\,3\cdot64\,+\,\frac{1}{4}\cdot256) \,- \,(0 \,-\,0\,+\,0)\right] \;= \;\L64\pi\)

 

[Guest]

oh that way does look easier, but we have to do this section by cross sections.... i guess i just made an algebraic error, which is good, so i actually know what im doing, thanks! yayy lol