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 = √2x + 1 , 0 ≤ x ≤ 3 ; 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. Compute the surface area obtained by revolving the curve Y is equal to the square root of 2 x minus 1 or 2 is less than or equal to x and x is equal to 8 about the x axis. OK. So it appears for this particular prompt or ultimately asked to compute or to solve for the surface area that is obtained by revolving the curvey. 42 is less than or equal to X and X is less than or equal to 8 about the X axis. So now that we know that we're ultimately trying to solve for this value of the surface area. Our first step that we need to take is we need to recall, based on our prior knowledge that the surface area generated by revolving a curve about the x axis is given by the following formula, which states that s, the surface area, is equal to the integral. Evaluated from A to B of 2 multiplied by pi multiplied by y multiplied by the square root of 1 plus Dy by dx in parentheses to the power of 2. DX fantastic. So now that we have our surface area formula, our next step now is we need to find Dy by DX. So DY by DX will be equal to D by Dx of parentheses of the square root of. 2 x minus 1 and this value for y is given to us for the curvey by the prom itself. So with that said, we will find that this will be equal to 1 divided by 2 multiplied by the square root of 2 x minus 1. Multiplied by d by dx of 2 x minus 1, which will be equal to, when we write our answer in its most simplified form, 1 divided by the square root of 2 x minus 1. Fantastic. So, with that said, we could therefore say that the parentheses of Dy by Dx to the 2 will be equal to 1 divided by 2 x minus 1. Fantastic. So now we need to substitute our new variables into our formula. So we'll find that the surface areas. is equal to the integral evaluated from 22/8 of 2 pi multiplied by the square root of 2 x minus 1 multiplied by the square root of 1 + 1 divided by 2 x minus 1. DX fantastic. So now we need to do a little bit of simplifying, so we need to simplify our current expression using a little bit of algebra, and when we do, we will find that the surface areas will be equal to 2 pi multiplied by the integral of 2, or specifically 2 pi multiplied by the integral evaluated from 2. 2/8 of the square root of 2 x minus 1 multiplied by the square root of 2 x divided by 2 x minus 1 dx and this is our expression written in its most simplified form. So, with that said, we now need to focus our attention on what the domain of, so once again, we need to figure out what the domain is for. So the domain of the square root of 2 x minus 1, and the domain D is square bracket 1/2 positive infinity. And now we need to figure out what the domain of the square root of 2X divided by 2X minus 1, and we're trying to find what the domain is for this for this particular expression. So with that said, we need to note that 2 x divided by 2X minus 1 is greater than or equal to 0, with 2 X minus 1 cannot equal 0. And now we need to figure out what our critical points are, so we need to recall a note that the fraction is 0 when the numerator is 0, so therefore 2 x is equal to 0 when X is equal to 0. And the fraction will be undefined when the denominator is zero, so therefore 2 X is equal to 1 will mean that X is equal to 1/2. So that will mean 4, so let's make a note that 4 X is less than 0, that will mean that 2 X divided by 2 X minus 1 will be greater than 0, which will mean that it's allowed, so I'll just put a capital A in quotes to mean that it's allowed. And for when 0 is less than X and X is less than 1/2, that will mean that 2 X divided by 2X minus 1 will be less than 0, which is not allowed, which I'll call NA for short. And finally, for when X is greater than 1/2, that will mean that 2 x divided by 2X minus 1 will be greater than 0, and this is allowed. So we need to make sure that we include X is equal to 0, since it makes the fraction 0. So therefore, the domain 4, so the domain for the square root of 2 X divided by 2X minus 1. Is parentheses infinity, 0 square bracket. Union parentheses 1/2 comma. Infinity in parentheses and since the functions, the square root of 2 x minus 1 and the square root of 2 x divided by 2 x minus 1 are both valid on square brackets of which square brackets indicate a closed interval. So the closed interval 2.8, that will mean the integral will become drum roll. So the surface areas will be equal to 2 pi multiplied by the integral evaluated from 22/8 of the square root of 2x multiplied by parentheses of 2x minus 1 divided by 2X minus 1DX. Fantastic. Which we could write as S is equal to 2 pi multiplied by the integral evaluated from 2 to 8 of the square root of 2 x dx. Fantastic. Which is S is equal to 2 multiplied by the, so it's gonna be 2 pi multiplied by the square root of 2 multiplied by the interval evaluated from 2 to 8 of the square root of XDX. Which is equal to S is equal to 2 pi multiplied by the square root of 2, multiplied by the integral, evaluated from 22/8 of x to the power of 1/22 DX. OK, we are on a roll here. So That will mean that s is equal to 2 pi multiplied by the square root of 2 multiplied by x to the power of 3 divided by 2 divided by 3 divided by 2, evaluated from 2 to 8. Which will be equal to 4 multiplied by pi multiplied by the square root of 2, divided by 3, multiplied by x multiplied by the square root of x, evaluated from 2 to 8. Which will mean when we begin our, so we can do. We need to do one more simplification first before we perform our evaluation, so. They actually we could do our evaluation now. So the surface areas is equal to 4 multiplied by pi, multiplied by the square root of 2 divided by 3, multiplied by square brackets of 8 multiplied by the square root of 8 minus 2 multiplied by the square root of 2. And when we write our answer in its most simplified form, we will find that it's equal to 112 pi divided by 3. And don't forget that our units are square units, since we're dealing with surface area. And that's it. That is our final answer. Hooray, we did it. So going back to the top to recap what we've learned, we could safely say that our final answer for the surface area. Is once again, 112 pi divided by 3 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 = √2x + 1 , 0 ≤ x ≤ 3 ; x-axis"

Watch next

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 = √2x + 1 , 0 ≤ x ≤ 3 ; x-axis"