# Thread: Centroid & Volume of region

1. ## Centroid & Volume of region

The region in the first quadrant bounded by the graphs of $y=x$ and $y=\frac{x^2}{2}$ is rotated around the line $y=x$ Find the centroid of the region and the volume of the solid of revolution.

2. Rotate the coordinate axes $\pi/4$.
The transformation will make a new graph.
Now simplify find the volume of the region around the x-axis.
(I will do this problem latter. Right now I am using one of my many computers. This one does not have a gprahing program.)
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You have,
$2y=x^2$ this describes your curve.
From conics you should be familar that a rotation of $\pi/4$ transforms,
$(x,y)\to (\cos \pi/4\cdot x-\sin \pi/4\cdot y,\sin \pi/4 \cdot x+\cos \pi/4 \cdot y)$
Equivalently,
$(x,y)\to \left( \frac{\sqrt{2}}{2} x-\frac{\sqrt{2}}{2} y, \frac{\sqrt{2}}{2} x+\frac{\sqrt{2}}{2}y\right)$
Thus,
$2\left( \frac{\sqrt{2}}{2} x+\frac{\sqrt{2}}{2}y\right)=\left( \frac{\sqrt{2}}{2} x-\frac{\sqrt{2}}{2} y\right)^2$
Thus,
$\sqrt{2} x+\sqrt{2}y=\frac{1}{2}x^2-\frac{1}{2} xy+\frac{1}{2}y^2$
$2\sqrt{2} x+2\sqrt{2} y=x^2-xy+y^2$
Now the problem is that this curve is not a function (if I only had a graph) and the shell formula needs to curve to be a function. So I am going to divide this curve into two curves that are functions and add their rotational volume about the y-axis. In order to do that I will first need to bring this to the form $y=f(x)$. To do that I need to solve for $y$.
First rewrite as,
$y^2-y(x+2\sqrt{2})+(x^2-2\sqrt{2} x)=0$
Thus,
$y=\frac{-x-2\sqrt{2}\pm \sqrt{x^2+4x\sqrt{2}+8-4x^2+8x\sqrt{2}}}{2}$
(TO BE CONTINUED)

3. I hope everyone had a nice holiday weekend, at least those stateside anyway. Just wanted to bump this to the top in hopes of the continuation....

4. Please, do not make me.
It would be a mess.
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Why do you even need it?
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5. Well, since you asked. I am taking a cracker-jack on-line Calc I class. This is the last question on an open-book test. I have answered the others, some with the help of this forum. I am averaging a whopping 72% in the class, and passing is all I care about right now Anyway, I can't really afford to leave this one blank, the adjunct usually gives partial credit for at least getting part of it right!

6. Originally Posted by Zeppelin
Well, since you asked. I am taking a cracker-jack on-line Calc I class. This is the last question on an open-book test. I have answered the others, some with the help of this forum. I am averaging a whopping 72% in the class, and passing is all I care about right now Anyway, I can't really afford to leave this one blank, the adjunct usually gives partial credit for at least getting part of it right!
Who to, ImPerfectHacker?

RonL