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

AI Thread Summary
The discussion focuses on calculating the volume of a solid formed by revolving the region bounded by the graphs of y=x and y=x^2 about the line y=x. The provided solution utilizes a specific formula for solids of revolution about an oblique axis, leading to the determination of the volume as pi/(30sqrt(2)). The integration process involves finding the limits of integration and expanding the integrand. The final calculation confirms the volume matches the book's answer. This method illustrates the application of calculus in determining volumes of solids of revolution.
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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.
 
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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:

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

We first find $x_i$ and $x_f$:

$$x^2=x$$

$$x(x-1)=0$$

Hence:

$x_i=0,\,x_f=1$

We are given:

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

Hence:

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

Expanding the integrand, we have:

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

Applying the FTOC, we find:

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

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