69-72. Volumes of solids Find the volume of the following soli...

Question

69-72. Volumes of solids Find the volume of the following solids.

70. The region bounded by y = 1/[x²(x² + 2)²], y = 0, x = 1, and x = 2 is revolved about the y-axis.

Accepted Answer

Hello there. Today we're gonna 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 generated when the region bounded by Y is equal to 1 divided by X multiplied by parenthesis of X2 + 12. The X axis Y is equal to 0, and the vertical lines X is equal to 1 and X is equal to 3 is revolved around the Y axis. Awesome. So it appears for this particular prompt we're ultimately trying to determine the volume of the solid that is generated when the region bounded to this curve of Y and the X axis Y is equal to 0 and the vertical lines X is equal to 1 and X is equal to 3. is revolved about the Y axis. Awesome. So with that in mind, let's take a moment to look at the figure that is provided to us for the prom itself, and we have our A general XY coordinate system where we have our vertical axis is the Y axis. And as you can see, we have an arrow indicating that we're revolving around the y axis, and we can see that on our Y axis we have it from -1 up to 1. And for our x-axis, the horizontal axis, it starts from the origin 00 and goes to 4. Then we have our two vertical lines or vertical curves we could say, at X equals 1 and X equals 3. And then we have our blue curve, which represents our curve for Y. So the blue curve is for Y is equal to 1 divided by X multiplied by parenthesis, X2 + 12. Fantastic. And we can see inside of our two vertical lines, we can see we have a shaded region. Under our curve. And we're going to use that to help us determine the volume. So now that we have a good visualization of what's happening for this particular problem, our first step that we need to take is we need to take a moment to understand the problem and what method we will need to use in order to solve this problem. So the region is between X is equal to 1 and X is equal to 3 above the x-axis. So I'll put an arrow next to x axis, so we know that it's above the X-axis. So with that in mind, we will now need to revolt this around the y axis. So that means the method that we need to recall how to use is that we need to note. That since we revolve around the y axis, and our function is given as, it's important to note that it's given as Y is equal to F of X, where F of X denotes a function. We can recall and use the shell method. So with that in mind, our second step is we need to use the shell method, and the shell method formula for a revolution around the y axis states that V, the volume, is equal to 2 pi multiplied by the interval from A to B of X multiplied by YD X. Where in this case, A will be equal to 1 and B will be equal to 3, and we will note that Y will be equal to 1 divided by X multiplied by parenthesis of X2 + 1 to 2. So we can go ahead and write that the volume V will be equal to 2 pi multiplied by the interval from 1 to 3 of X multiplied by 1 divided by X multiplied by parenthesis of X2 + 1 to 2DX which will be equal to 2 pi multiplied by the interval from 1 to 3 of 1 divided by X to the power of 2. Plus 1 in parenthesis to the power of 2 DX. fantastic. The X in the numerator, so this X in the numerator. And denominator cancels out nicely. So with that in mind, our third step is we need to simplify our integral. So the volume interval will reduce to V is equal to 2 pi multiplied by the integral from 1 to 3 of 1 divided by parenthesis, X2 + 12 D X. And this is because the X in the numerator denominator will cancel out nicely. So with that in mind, our next step is we need to, this is going to be our step number 4, is we need to evaluate the integral. So in order to do that, we need to recall and use the standard interval formula. So we need to note that the interval of DX divided by parentheses X 2 plus a 2 in 2. Will be equal to x divided by 2A 2 multiplied by parenthesis of X 2 plus a 2. Plus 1 divided by 2 multiplied by 8 to the power of 3 multiplied by inverse tangent of X divided by A plus C, where C denotes some constant value. And we need to note, and let's note this in blue, that A is equal to 1. So because A is equal to 1, we can go ahead and write that the integral of DX divided by parentheses X 2 + 12 will be equal to x divided by 2 multiplied by 12 multiplied by parentheses X 2 + 1. Plus 1 divided by 2 multiplied by 1 to the power of 3 multiplied by inverse tangent of X plus C, which will be equal to x divided by 2 multiplied by parenthesis of X2 + 1. Plus 1/2 or 1 divided by 2 multiplied by the inverse tangent of X plus C. Fantastic. So our 5th step now is we need to compute the definite interval. So we need to take the interval from 1 to 3 of, so we're going to take the integral evaluated from 1 to 3 of 1 divided by parentheses X to the power of 2. Plus 1 to 2 DX which will be equal to square brackets of X divided by 2 multiplied by parenthesis of X to the power 2 + 1 + 1/2 multiplied by. Arc tangent. Of X evaluated from 1 to 3, so we now need to calculate each term, so we need to note that at X equals 3 we can state that once again we're plugging in a value of 3 in for X, so we can write that 3 divided by 2 multiplied by 9 + 1 + 1/2. Multiplied by inverse tangent of 3, so arc tangent is just another way of writing inverse tangent, potato potato. So that will mean, so once again we're taking 3 divided by 2 multiplied by 9 + 1 plus 1/2 multiplied by inverse tangent of 3, which will be equal to 3 divided by 20 plus 1 divided by 2 multiplied by inverse tangent of 3. Fantastic. And at X is equal to 1, so now we're plugging in a value of 1 in for X. We will get 1 divided by 2 multiplied by parentheses 1 + 1. Plus 1/2 multiplied by the inverse tangent of 1, which will be equal to 1/4 plus 1/2 multiplied by pi divided by 4, which will be equal to 1/4 plus pi divided by 8. Fantastic. Our 6th step is we need to calculate the definite interval value. So when we do that, we will find that the integral evaluated from 1 to 3 of 1 divided by parentheses X to the power to. Plus 1 in parentheses to the power of 2 D X will be equal to parentheses 3 divided by 20 plus 1 divided by 2 multiplied by inverse tangent of 3. Minus parentheses 1 divided by 4 plus pi divided by 8. Fantastic, which will be equal to drum roll, 3 divided by 20 minus 1/4, so 3 divided by 20, minus 1 divided by 4 plus 1 divided by 2 multiplied by inverse tangent of 3 minus pi divided by 8. Fantastic. So now we need to simplify our constants, and when we do, we will find that 3 divided by 20 minus 1 divided by 4 will be equal to 3 divided by 20 minus 5 divided by 20, which will be equal to and we simplified down to its most simple form, -1 divided by 10. So that will mean that we can go ahead and write that the interval evaluated from 1 to 3 of 1 divided by parenthesis of X2 + 12. The X will be equal to -1 divided by 10 plus 1 divided by 2 multiplied by inverse tangent of 3 minus pi divided by 8. Fantastic. So that will mean that our 7th step is we need to solve for the final volume formula. So we need to multiply by pi. So 2 pi specifically. So we need to multiply by 2 pi, and when we do, we will find that the volume V is equal to 2 pi multiplied by parentheses -1 divided by 10 plus 1 divided by 2 multiplied by inverse tangent of 3 minus pi divided by 8. Fantastic. And when we do that, we will find that it's equal to drum roll, 2 pi multiplied by parenthesis negative 1 divided by 10 plus 2 pi multiplied by 1 divided by 2 multiplied by inverse tangent of 3 minus 2 multiplied by pi divided by 8. Fantastic. And when we simplify, we will find that V is equal to -2 divided by 10 plus pi multiplied by inverse tangent of 3 minus 2 divided by 4, which will be equal to negative pi divided by 5 plus pi multiplied by inverse tangent of 3 minus 2 divided by 4. And that's it. We've officially solved for this problem, but writing it in a more neat fashion, you could write our final answer as V the volume is equal to pi multiplied by inverse tangent of 3 minus pi divided by 5 minus pi to 2 divided by 4, and that is our final answer. Hooray, we did it. So it's sometimes some professors or teachers do not like it when you write a negative value first at the beginning of your expression. So it's sometimes it's better if you rearrange your expression to make sure that you don't have a negative symbol at the beginning. So with that in mind, to recap what we've learned, that will mean that looking at the prom itself, our final answer, without a doubt, will have to be once again. Drum roll. So pie. Multiplied by inverse tangent of 3 plus pi divided by 5 minus pi to 2 divided by 4, 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.

69-72. Volumes of solids Find the volume of the following solids.
70. The region bounded by y = 1/[x²(x² + 2)²], y = 0, x = 1, and x = 2 is revolved about the y-axis.

Key Concepts

Volume of Solids of Revolution

Shell Method

Setting up Proper Integral Limits and Functions

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