Volume of solid of rotation: y = x, y = x^2/2 about y = x

[confused_07]

The region in the first quadrant bounded by the graphs of y = x and y = (x^2/2) is rotated around the line y = x. Find the volume.

V = int [c,d] pi [ x(right)^2 - x(left)^2] dy

I know when you rotate around a number on the x axis, say 1, then you add 1 to the right and left before your square. What do you add for this problem? Would you add 'x'?

 

[jacket81]

Okay, I could be wrong here.
But finding the equation of it inverted and then rotating it 45 degrees clockwise, I get equation y=squareroot(((sqr(2x)-x)^2)+((x^2/2)-x)^2).

Then, the volume is pi times the intergral of that squared from x=2 to x=0.

 

[tkhunny]

Well, you have to use what you know.

Hint: \(\displaystyle \theta\;=\;\frac{\pi}{4}\)

\(\displaystyle \cos{\theta}\;=\;\frac{\sqrt{2}}{2}\)

\(\displaystyle \sin{\theta}\;=\;\frac{\sqrt{2}}{2}\)

\(\displaystyle x\;=\;\frac{\sqrt{2}}{2}(u-v)\)

\(\displaystyle y\;=\;\frac{\sqrt{2}}{2}(u+v)\)

Consider rotating the function \(\displaystyle v\;=\;u\;+\;\sqrt{2}\;-\;\sqrt{4u\sqrt{2}+2}\) from \(\displaystyle u\;=\;0\) to \(\displaystyle u\;=\;2\sqrt{2}\) around the line v = 0.

I get \(\displaystyle \frac{2}{15}\sqrt{2}\pi\), but I'm willing to be wrong.

If you don't like the problem you have, try a different problem!

Having said that, there is probably a really convenient way. I don't happen to know it off the top of my head. You discussed nothing helpful in class?

 

[confused_07]

I don't have a class. This is a distance learning course (ie teach yourself), hence why I ask alot of questions. I could email my instructor, but I get faster responses here.

 

[tkhunny]

What unit are you in. I suppose it has a title or section heading.

 

[confused_07]

Are you talking about the Chapter of the text? I just finished "Centroids of plane regions and curves", which is part of the chapter (Applications of the integral)

 

[pka]

I don’t think axis-rotation has anything to do with this problem.
Have you had (i.e. is it in your text material) Theorem of Pappus?
Because you have found the centroid of that region and the region is entirely on one side of the line y=x you can apply the theorem.

 

[tkhunny]

That does make it easier. See, I told you there was a convenient way.

The rule of doctors, "If you don't think of it, you won't diagnose it."

 

[galactus]

This is a toughy compared to most 'solid of revolution problems, but not unheard of.

The area of the region is:

\(\displaystyle \L\\A=\int_{0}^{2}\int_{\frac{x^{2}}{2}}^{x}dydx=\frac{2}{3}\)

Now, that centroid thing you mentioned:

\(\displaystyle \L\\\bar{x}=\frac{3}{2}\int_{0}^{2}\int_{\frac{x^{2}}{2}}^{x}xdydx=1\)

\(\displaystyle \L\\\bar{y}=\frac{3}{2}\int_{0}^{2}\int_{\frac{x^{2}}{2}}^{x}ydydx=\frac{4}{5}\)

Now, we need the shortest distance from the centroid we found to the axis we're revolving it around. Ol' Pythagoras:

\(\displaystyle \L\\D(x)=(x-1)^{2}+(x-\frac{4}{5})^{2}\)

\(\displaystyle \L\\D'(x)=4x-\frac{18}{5}=0, \;\ x=\frac{9}{10}\)

\(\displaystyle \L\\D(9/10)=\frac{1}{50}\)

\(\displaystyle \L\\\sqrt{D}=\sqrt{\frac{1}{50}}=\frac{\sqrt{2}}{10}\)

The volume is \(\displaystyle \L\\2{\pi}(\frac{\sqrt{2}}{10})(\frac{2}{3})=\frac{2\sqrt{2}{\pi}}{15}\)

Same as TKH. I reckon it's correct then.

 

[jacket81]

Sorry, ignore my post that i sent. I made a mistake in my calculations, sorry.

 

[tkhunny]

I love it when:

1) Different methods produce the same, unique result,
2) 2000 year old guys are smarter than I am.