Areas of Surfaces of Revolution
In Exercises 23–26, find...

Question

Areas of Surfaces of Revolution

In Exercises 23–26, find the areas of the surfaces generated by revolving the curves about the given axes.

_______

y = √4y ― y² , 1 ≤ y ≤ 2 ; y-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 area of the surface generated by revolving the curve x is equal to the square root of 9 y minus y to the 2. Or 2 is less than or equal to y and y is less than or equal to 5 about the y axis. OK, so it appears for this particular prompt for us to ultimately determine or find out what the area of the surface is that is once again generated by revolving the curve. Or revolving this curve X about the Y axis, provided we have an interval 2 is less than or equal to Y and Y is equal to 5. OK, so now that we know that we're ultimately trying to determine what this. Area of the surfaces. Our first step that we need to take is based on our prior knowledge. We need to recall and note that the surface area that is generated by revolving a curve about the y axis is given as, so the formula is given as. S, the surface area is equal to the integral evaluated from A to B of 2 pi multiplied by x multiplied by the square root of 1 + parentheses of dx by Dy to the power of 2. See why So now, now that we know our formula for the surface area, we now need to find the value of DX by DY. So we need to solve for DX by DY and DX by DY will be equal. To D by Dy of parentheses, so we are told that our curve is x is equal to the square root of 9Y minus y to the 2. So we need to take Dy, so D by Dy of. The square root of 9 y minus y to the power of 2, which will be equal to 1 divided by 2 multiplied by the square root of 9 y minus y to the power of 2 multiplied by d by dy of 9 y minus y to the power of 2. Which will be all equal to 9 minus 2y divided by 2 multiplied by the square root of 9 y minus y to the power of 2. That will therefore mean that parentheses of d x by Dy all to the power of 2 will be equal to parentheses of 9. -2 y divided by 2 multiplied by the square root of 9 y minus y to the power of 2 to the power of 2, which will be equal to when we write our final answer in its most simplified form, 81 minus 36 y. Plus 4 multiplied by y to the power of 2, all divided by 36 y minus 4 multiplied by y to the power of 2. So now we need to substitute this value back into our surface area formula. So as the surface area is equal to the integral, evaluate from 2 to 5 of 2 pi multiplied by the square root of 9 Y minus y to the 2 multiplied by the square root of 1 + 81 minus 36 Y. Plus Or multiplied by y to the power of 2. All divided by 36 Y. Minus -4y to the 2. Fantastic. So, and don't forget DY. Fantastic. So moving right along here, when we simplify our expression down to its most simple form, we will find, I'm going to skip a few steps, so you just use a little bit of algebra, but you'll find that the surface areas will be equal to 2 pi multiplied by the integral. from 2 to 5 of the square root of 9 y minus y to the power 2 multiplied by the square root of 81 divided by 4 multiplied by parentheses of 9 y minus y to the power of 2. Do you why? Fantastic. And let us note now that the domain which I'll call d, so the domain of the square root of 9 y minus y to the power of 2. Is, so once again our domain is square brackets of 0.9, which means the closed interval 0.9, means that's what square brackets represent. And the domain of the square root of 81 divided by 4 multiplied by parentheses of 9 y minus y to the power of 2. Is parentheses of 0.9, which parentheses represent an open interval. So with that in mind, we can safely say that both functions are valid, which I'll call V for short. So both functions are valid on 2 is less than or equal to y and y is less than or equal to 5. Thus we can go ahead and write that the surface areas is equal to 2 pi multiplied by the integral evaluated from 2 to 5 of the square root of 81 multiplied by parentheses of 9 y minus y to the power of 2. Divided by. 4 multiplied by 9, so it's gonna be 4 multiplied by parentheses of 9 Y minus y to the power of 2. DY Which will be equal to, or we could write that S is equal to 2 pi multiplied by the integral evaluated from 2 to 5 of 9 divided by 2 DY, which will mean that S is equal to 9 pi multiplied by the integral evaluated from 2 to 5 of DY. Which will mean that s is equal to 9 pi multiplied by y evaluated from 2 to 5, which will mean that s is equal to 9 pi multiplied by parentheses of 5 minus 2. Which will be equal to when we write our final answer in its most simplified form 27 pi and don't forget, since we're solving for the for the surface area, that means our units will be square units, and that's it, we've officially solved for our final answer. Hooray, we did it. So, looking at the prom itself, we could safely say once again that our final answer for the surface area is, once again, 27 pi square 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.

Areas of Surfaces of RevolutionIn Exercises 23–26, find the areas of the surfaces generated by revolving the curves about the given axes. _______ y = √4y ― y² , 1 ≤ y ≤ 2 ; y-axis

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Areas of Surfaces of Revolution
In Exercises 23–26, find the areas of the surfaces generated by revolving the curves about the given axes.
_______
y = √4y ― y² , 1 ≤ y ≤ 2 ; y-axis