Volumes
Find the volume of the solid generated by revolv...

Question

Volumes

Find the volume of the solid generated by revolving the “triangular” region bounded by the curve y = 4/x³ and the lines x = 1 and y = 1/2 about

a. the x-axis

Accepted Answer

Hello there. Today we are going 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 the curve. Y is equal to 9 divided by x to the power of 2. The vertical line x is equal to 1, and the horizontal line y is equal to 1 about the x axis. OK. So it appears for this particular prompt we're asked to find the volume of the solid based on all the information that is provided to us by the prom itself. So now that we know that we're ultimately trying to solve for the volume of this particular solid. We need to know, based on the information that is provided to us by the prom itself, we need to be able to recognize, based on our prior knowledge that we can recall and use the washer method to solve for this particular problem. And as we should recall, the general formula for the volume when rotating about the x axis is written as the volume V is equal to pi multiplied by the integral evaluated from A to B of parentheses of square brackets of R X. To the power of 2 minus square brackets of lowercase r x to the power of 2 dx. OK, where we need to note that capital R X is the outer radius and lowercase r x is the inner radius. So now, at this stage, we need to find the x bounds by intersecting Y is equal to 9 divided by x 2. And Y is equal to 1. So that will mean that 9 divided by X 2 will be equal to 1, which will mean that X 2 is equal to 9, which will mean that X is equal to plus or minus 3. And we need to take that X is greater than 0 because the region in the problem is bounded by the vertical line. At X is equal to 1. And also X is equal to 3, so we need to note that X is equal to 1 and X is equal to 3. Fantastic. So, moving right along here. Therefore, we could say that the region is for 1 is less than or equal to X and X is less than or equal to 3. Fantastic. So now at this point, we need to consider our radii, which are the distances from the x axis. So we need to note that based on our prior knowledge that capital R X will be equal to 9 divided by x 2. And this is for capital R X, and lowercase r x will be equal to 1. And we need to note, so let's make a note in blue, that capital R X is greater than or equal to lowercase r x on the closed interval, square brackets of 1.3. So with this in mind, we can now set up and evaluate our integral and start to solve for the volume of this solid. So we could go ahead and write that the volume V is equal to pi multiplied by the integral evaluated from 1 to 39 divided by x 2. To the 2 minus 12. DX Which will mean that V is equal to pi multiplied by the integral evaluated from 1 to 3 ses of 81 divided by x 4. -1 DX. Which we could go ahead and write that V is equal to pi multiplied by the integral evaluated from 1 to 3 of parentheses. 81 multiplied by x to the power of -4 minus 1 dx. Which will go ahead and write that V is equal to pi multiplied by square brackets of 81 divided by -3 multiplied by x to the power of -3 minus x evaluated from 1 to 3. So now we could start to slowly begin our evaluation. So you go ahead and write that V is equal to negative, so V is equal to pi multiplied by square brackets of 27 divided by x to the power of 3. Plus X. In square brackets evaluated from 1 to 3. Which we could go even further and write that V is equal to negative pi multiplied by square brackets parentheses 27 divided by x to the, so it's gonna be 27 divided by. 3 to the 3 because now we need to plug in our evaluation variables. So it's going to be 27 divided by 3 to the power of 3 plus 3 in parentheses minus parentheses of 27 divided by 1 to 3 plus 1. Which will mean that when we plug in this expression into our calculator, we will find that V is simply equal to pi multiplied by square brackets of 4 minus 28. Which will mean that our final answer for the volume will be V is equal to 24 pi, and don't forget that our units will be cubic units because we're dealing with volume. And that's it. We've officially solved for this problem. So going back to look at the prom itself, we could safely say without a doubt that the volume of this particular solid is once again, 24 pi cubic units, and that's it. That is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

VolumesFind the volume of the solid generated by revolving the “triangular” region bounded by the curve y = 4/x³ and the lines x = 1 and y = 1/2 abouta. the x-axis

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Volumes
Find the volume of the solid generated by revolving the “triangular” region bounded by the curve y = 4/x³ and the lines x = 1 and y = 1/2 about
a. the x-axis