Bob's question at Yahoo Answers regarding mimimizing a solid of revolution

In summary, the question is asking for the minimum volume of a solid created by rotating the function f(x)=p/x^(p) around the x-axis over the interval [1,infinity). Using the disk method, we find that the minimum volume occurs when p=1.
  • #1
MarkFL
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Here is the question:

Solid of Revolution Calculus 2 question.?

For a positive real number p, define f(x)=p/x^(p). Find the minimum value of the volume of the solid created by rotating this function around the x-axis over the interval [1,infinity).

I have posted a link there to this thread so the OP can view my work.
 
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  • #2
Hello bob,

First, let's write the given function as:

\(\displaystyle f(x)=px^{-p}\)

Now, using the disk method, we find that the volume of an arbitrary disk is:

\(\displaystyle dV=\pi r^2\,dx\)

where:

\(\displaystyle r=f(x)=px^{-p}\)

And so we have:

\(\displaystyle dV=\pi p^2 x^{-2p}\,dx\)

Summing us all of the disks, we may state:

\(\displaystyle V=\pi p^2\int_1^{\infty}x^{-2p}\,dx\)

This is an improper integral with the unbounded upper limit, so we may write:

\(\displaystyle V=\pi p^2\lim_{t\to\infty}\left(\int_1^{\infty}x^{-2p}\,dx \right)\)

Applying the FTOC, we have:

\(\displaystyle V=\pi p^2\lim_{t\to\infty}\left(\left[\frac{x^{1-2p}}{1-2p} \right]_1^{t} \right)\)

\(\displaystyle V=\frac{\pi p^2}{1-2p}\lim_{t\to\infty}\left(\frac{1}{t^{2p-1}}-1 \right)\)

For \(\displaystyle \frac{1}{2}<p\) we have:

\(\displaystyle V(p)=\frac{\pi p^2}{2p-1}\)

To determine the critical value(s), we need to differentiate with respect to $p$ and equate the result to zero. Using the quotient rule, we find:

\(\displaystyle V'(p)=\frac{(2p-1)(2\pi p)-\left(\pi p^2 \right)(2)}{(2p-1)^2}=\frac{2\pi p(p-1)}{(2p-1)^2}\)

For \(\displaystyle \frac{1}{2}<p\) we have:

\(\displaystyle p=1\)

If we observe that for \(\displaystyle \frac{1}{2}<p<1\) we have $V'(p)<0$ and for $1<p$ we have $V'(p)>0$, we may therefore conclude that this critical value is at a minimum.

Thus, we may conclude that the described solid of revolution is minimized when $p=1$.
 

FAQ: Bob's question at Yahoo Answers regarding mimimizing a solid of revolution

1. How do you minimize a solid of revolution?

To minimize a solid of revolution, you need to find the critical points of the function representing the solid. These points will be the locations where the derivative of the function is equal to zero or undefined. Then, use the second derivative test to determine if the critical points are maximum, minimum, or inflection points. The minimum point will be the smallest value of the solid.

2. What is a solid of revolution?

A solid of revolution is a three-dimensional shape formed by rotating a two-dimensional shape around an axis. The resulting shape is a solid with a circular cross-section.

3. What is the formula for finding the volume of a solid of revolution?

The formula for finding the volume of a solid of revolution is V = π∫abf(x)2dx, where a and b are the limits of integration and f(x) is the function representing the solid.

4. How do you find the critical points of a function?

To find the critical points of a function, you need to take the derivative of the function and set it equal to zero. Then, solve for the variable to find the critical points. You may also need to check for points where the derivative is undefined.

5. What is the second derivative test?

The second derivative test is a method used to determine if a critical point of a function is a maximum, minimum, or inflection point. To use this test, you need to take the second derivative of the function and evaluate it at the critical point. If the second derivative is positive, the critical point is a minimum. If the second derivative is negative, the critical point is a maximum. If the second derivative is zero, the test is inconclusive.

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