65-76. Volumes Find the volume of the described solid of revol...

Question

65-76. Volumes Find the volume of the described solid of revolution or state that it does not exist.

66. The region bounded by f(x) = (x^2 + 1)^(-1/2) and the x-axis on the interval [2, ∞) is revolved about the x-axis.

Accepted Answer

Hello there. Today we want to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Find the volume of the solid obtained by revolving the region bounded by F of X is equal to parenthesis X2 + 16 to the power of negative 1/2. And the x axis on the interval square bracket 0. infinity parenthesis about the x axis. Awesome. So it appears for this particular problem we're ultimately trying to determine what the volume is of this particular solid. Using the information provided to us by the prom itself. So now that we know that we're ultimately trying to solve for the volume, let's read off our multiple choice answers to see what our final answer might be, noting that they're all in the same units of cubic units. So A is pi to the power 2 divided by 6, B is pi to power 2 divided by 4, C is pi to power 2 divided by 8, and finally D is pi divided by 8. Fantastic. So our first step that we need to take in order to solve this particular problem is we need to recall and use the disk method. So with that in mind, we need to recall and note that the volume of revolution about the X axis is given by the volume is equal to pi multiplied by the integral from 0 to infinity of square brackets of F of X. So the power of 2 DX fantastic, and we also need to note that we are given that F of X to the power of 2 is equal to parentheses 1 divided by the square root of X2 + 16 to 2, which is equal to 1 divided by X2 + 16. So therefore we can go ahead and write that the volume V is equal to pi multiplied by the integral from 0 to infinity of 1 divided by X to the power of 2 + 16 d X. Fantastic. Our second step now is we need to recall and use a standard integral. So what we need to do, since we're recalling how to use a standard interval, we need to recall and use the following formula that states that the integral. Of 1 divided by X 2 plus a power of 2 D X is equal to 1 divided by a multiplied by inverse tangent of X divided by a plus C, where C represents some constant. We also need to recall and note that for our case, A to the power of 2 is equal to 16, so that will mean that A is equal to 4. So we can go ahead and write that the integral from 0 to infinity of 1 divided by X to the power of 2. Plus 16 DX is equal to square brackets of square brackets of 1 divided by 4 multiplied by inverse tangent of X divided by 4. Evaluated from 0 to infinity. Exactly. So that now, so now what we're going to be doing is we're going to be evaluating our bounds. So at this point, since we're evaluating our bounds, we need to note that as X approaches positive infinity, so this is as X approaches positive infinity, that will mean that inverse tangent of X divided by 4 will approach pi divided by 2. And we need to note that at X equals 0, that will mean that inverse tangent of 0 will equal 0. So with that in mind, therefore, the integral from 0 to infinity of 1 divided by X to the power 2 + 16 DX will be equal to 1/4 multiplied by pi divided by 2, which will be equal to pi divided by 8. So now, our 3rd and final step. So once again, this is our last step. So our third step is the last step, and we need to multiply by pi now. So V is equal to pi, multiplied by pi divided by 8, which will be equal to pi to power 2 divided by 8. And that is our final answer. Hooray, we did it. So, our final answer without a doubt is pi to the power 2 divided by 8, and don't forget that our units are cubic units. And that's it. We've officially solved for this problem. So looking at our multiple choice answers, the correct answer without a doubt will have to be multiple choice answer letter C, which once again is divided by pi to the power 2 divided by 8 cubic units. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

65-76. Volumes Find the volume of the described solid of revolution or state that it does not exist.
66. The region bounded by f(x) = (x^2 + 1)^(-1/2) and the x-axis on the interval [2, ∞) is revolved about the x-axis.

Key Concepts

Volume of Solids of Revolution

Improper Integrals and Convergence

Function Behavior and Boundedness

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