Volume of Solid of Revolution about Oblique Axis - Kyle's Question

[MarkFL]

Here is the question:

Find the volume of the solid formed by revolving about y=x?

Find the volume of the solid formed by revolving the region bounded by the graphs of y=x and y=x^2 about the line y=x

Additional Details:

The book's given answer is pi/(30sqrt(2))

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Kyle,

Please see http://mathhelpboards.com/math-notes-49/solid-revolution-about-oblique-axis-rotation-6683.html for the development of the general formula:

\(\displaystyle V=\frac{\pi}{\left(m^2+1 \right)^{\frac{3}{2}}}\int_{x_i}^{x_f} \left(f(x)-mx-b \right)^2\left(1+mf'(x) \right)\,dx\)

We first find $x_i$ and $x_f$:

\(\displaystyle x^2=x\)

\(\displaystyle x(x-1)=0\)

Hence:

$x_i=0,\,x_f=1$

We are given:

$m=1,\,b=0,\,f(x)=x^2\implies f'(x)=2x$

Hence:

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

Expanding the integrand, we have:

\(\displaystyle V=\frac{\pi}{2\sqrt{2}}\int_{0}^{1} 2x^5-3x^4+x^2 \,dx\)

Applying the FTOC, we find:

\(\displaystyle V=\frac{\pi}{2\sqrt{2}}\left[\frac{1}{3}x^6-\frac{3}{5}x^5+\frac{1}{3}x^3 \right]_{0}^{1}\)

\(\displaystyle V=\frac{\pi}{2\sqrt{2}}\left(\frac{1}{3}-\frac{3}{5}+\frac{1}{3} \right)=\frac{\pi}{2\sqrt{2}}\left(\frac{1}{15} \right)=\frac{\pi}{30\sqrt{2}}\)