Question

Comparing volumes Let R be the region bounded by y=1/x^p and the x-axis on the interval [1, a], where p>0 and a>1 (see figure). Let Vₓ and Vᵧ be the volumes of the solids generated when R is revolved about the x- and y-axes, respectively.

d. Find a general expression for Vᵧ in terms of a and p. Note that p=2 is a special case. What is Vᵧ when p=2?

Accepted Answer

Identify the region R bounded by the curve $$y = \frac{1}{x^p}$$, the x-axis, and the vertical lines $$x=1$$ and $$x=a$$, where $$p \u003e 0$$ and $$a \u003e 1. $$Since the solid is generated by revolving the region R about the y-axis, use the method of cylindrical shells to find the volume $$V_y. $$The formula for the volume using cylindrical shells is:

$$V_y = 2\pi \int_{x=1}^{a} (	ext{radius})(	ext{height}) \, dx$$

where the radius is the distance from the y-axis to the shell (which is $$x$$), and the height is the function value $$y = \frac{1}{x^p}. $$Set up the integral for $$V_y as$$:

$$V_y = 2\pi \int_1^a x \cdot \frac{1}{x^p} \, dx = 2\pi \int_1^a x^{1-p} \, dx$$ Evaluate the integral $$\$$int_1$$^a x$$^{1-p}$$ \, dx. $$For $$p 
eq 2$$, use the power rule for integration:

$$\int x^{m} \, dx = \frac{x^{m+1}}{m+1} + C$$

where $$m = 1 - p. So$$,

$$\int_1^a x^{1-p} \, dx = \left[ \frac{x^{2-p}}{2-p} ight]_1^a = \frac{a^{2-p} - 1}{2-p}$$ Substitute this back into the expression for $$V_y to $$get the general formula:

$$V_y = 2\pi \cdot \frac{a^{2-p} - 1}{2-p}$$

For the special case when $$p=2$$, the integral becomes:

$$\int_1^a x^{-1} \, dx = \left[ \ln x ight]_1^a = \ln a

So$$,

$$V_y = 2\pi \ln a$$

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Key Concepts

Volume of Solids of Revolution about the y-axis

Shell Method Integration Setup

Handling Special Cases in Integration (p=2)