Volume of Region Between y=x and y=-x²+2x

[suluclac]

The region between y = x & y = -x² + 2x revolves around y = x. Determine the volume.

 

[MarkFL]

I get:

\(\displaystyle V=\frac{\pi}{2^{\frac{3}{2}}}\int_{0}^{1} \left(x-x^2 \right)^2\left(3-2x \right)\,dx=\frac{\pi}{2^{\frac{3}{2}}\cdot15}\)

 

[suluclac]

This problem is similar to this.
I tried using the theorem of Pappus. Is \(\displaystyle \pi/6\) incorrect?
How did you get your answer?

 

[MarkFL]

This problem is similar to this.
I tried using the theorem of Pappus. Is \(\displaystyle \pi/6\) incorrect?
How did you get your answer?

I used the formula I derived here:

http://mathhelpboards.com/math-notes-49/solid-revolution-about-oblique-axis-rotation-6683.html

Using that formula, I don't get the same answer you did, but I could have made a mistake applying the formula.

 

[suluclac]

Spoiler

They both intersect at x = 0, 1. Now find the center of mass.
\(\displaystyle M_x=\frac{1}{2}\int_0^1\left(x^4-4x^3+4x^2-x^2\right)\,dx=\frac{1}{2}\int_0^1\left(x^4-4x^3+3x^2\right)\,dx=\frac{1}{10}\)
\(\displaystyle M_y=\int_0^1x\left(-x^2+2x-x\right)\,dx=\int_0^1x\left(-x^2+x\right)\,dx=\int_0^1\left(-x^3+x^2\right)\,dx=\frac{1}{12}\)
\(\displaystyle A=M=\int_0^1\left(-x^2+2x-x\right)\,dx=\int_0^1\left(-x^2+x\right)\,dx=\frac{1}{6}\)
\(\displaystyle \left(\bar{x}\text{, }\bar{y}\right)=\left(\frac{M_y}{M}\text{, }\frac{M_x}{M}\right)=\left(\frac{1/12}{1/6}\text{, }\frac{1/10}{1/6}\right)=\left(\frac{1}{2}\text{, }\frac{3}{5}\right)\)
Distance from this point to y = x. 0 = x - y
\(\displaystyle d=\frac{|(1)(1/2)-(1)(3/5)|}{\sqrt{2}}=\frac{1}{10\sqrt{2}}\)
Theorem of Pappus
\(\displaystyle V=2\pi dA=2\pi\left(\frac{1}{10\sqrt{2}}\right)\left(\frac{1}{6}\right)=\frac{\pi}{30\sqrt{2}}\)