Volume of Solid of Revolution of f(x)

[Theia]

Let $$f(x) = x^3 + 4x^2 - x + 5$$ revolve about the line $$y(x) = -x + 5$$. There will form one solid with finite volume. Find the volume of that solid.

 

[MarkFL]

My solution:

Spoiler

In the following tutorial:

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

I developed the following formula:

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

To find the limits of integration, we need to find where the given cubic function, and the axis of rotation meet:

$$x^3+4x^2-x+5=5-x$$

$$x^2(x+4)=0$$

$$x=-4,0$$

Thus, we now have:

$$V=\frac{\pi}{2^{\frac{3}{2}}}\int_{-4}^{0} \left(x^3+4x^2\right)^2\left(2-3x^2-8x\right)\,dx$$

Expanding the integrand, we have:

$$V=\frac{\pi}{2^{\frac{3}{2}}}\int_{-4}^{0} -3x^8-32x^7-110x^6-112x^5+32x^4\,dx$$

Application of the FTOC gives us:

$$V=\frac{\pi}{2^{\frac{3}{2}}}\cdot\frac{32768}{105}=\frac{2^\frac{27}{2}\pi}{105}$$

 

[Theia]

Yeah, that's right. Well done! ^^