# Volume of Revolution about $y$-axis. Partial Fraction?

I need to find the volume of the solid generated by revolving, about the $y$-axis, the area between the $x$-axis and the curve $y=\frac{1}{x^2+8x+7}$ for $x \in [0,1]$. Also need to rotate about the line $x=-2$. Having trouble setting up the integrals and finding the limits of integration.

• Yes, partial fractions is the right way to go. $x^2+8x+7 = (x+1)(x+7)$ and $6 \cdot 7 = 42$ Commented Feb 26, 2018 at 11:50
• About the rotation around $x = -2$, think of a translation from the $y$ axis to the line $x = -2$ which is parallel to the $y$ axis. Commented Feb 26, 2018 at 11:53
• So once I do partial fraction decomposition how would the integral be set up? Commented Feb 26, 2018 at 12:07

Partial fraction decomposition won't be necessary to set up the integrals, but it will help when you want to evaluate them. For both rotations, since they involve rotating a function in terms of $$x$$ around a vertical line, it would be easiest to use the shell method to set up the integrals.

Imagine the first solid of revolution, obtained from rotating the graph of $$f(x)=\frac{1}{x^2+8x+7}$$ around the $$y$$-axis. If we split that object into cylindrical shells, the height of each shell at any point $$x\in[0,1]$$ will be given by $$f(x)$$, and the radius of each shell will just be $$x$$. Therefore, since we only care about the region $$0\le x\le1$$, and because the volume of a cylinder is given by $$V=2\pi rh$$, the integral will look like this:

$$V_1=2\pi\int_0^1\frac{x}{x^2+8x+7}\,\mathrm{d}x$$

Now, imagine the second solid of revolution, obtained from rotating the graph around the line $$x=-2$$. The height will again be given by $$f(x)$$, and because we're rotating the graph around a line $$2$$ units away from the left-most boundary of our region, the radius of any shell at a given point $$x\in[0,1]$$ will be $$x+2$$. That means our integral will look like this:

$$V_2=2\pi\int_0^1\frac{x+2}{x^2+8x+7}\,\mathrm{d}x$$

Pretty similar, right? You can now use partial fraction decomposition to simplify the integrals a little bit:

$$V_1=2\pi\int_0^1\frac{x}{(x+1)(x+7)}\,\mathrm{d}x=2\pi\int_0^1\frac{-\frac{1}{6}}{x+1}+\frac{\frac{7}{6}}{x+7}\,\mathrm{d}x$$

$$V_2=2\pi\int_0^1\frac{x+2}{(x+1)(x+7)}\,\mathrm{d}x=2\pi\int_0^1\frac{\frac{1}{6}}{x+1}+\frac{\frac{5}{6}}{x+7}\,\mathrm{d}x$$

These integrals are still pretty similar, and I trust you can figure out how to take it from here!

Hint: $$\int\frac{a}{x+b}\,\mathrm{d}x=a\ln|x+b|+C$$