Find the area of the surface generated when the given curve is...

Question

Find the area of the surface generated when the given curve is revolved about the given axis.

y=x^3/2−x^1/2 / 3, for 1≤x≤2; about the x-axis

Accepted Answer

Welcome back, everyone. In this problem, we want to find that the surface area generated by revolving the curve Y equals root X for X greater than or equal to 2, but less than or equal to 5 above the X axis. Now to help us figure this out, let's ask ourselves what do we know about the surface area generated by revolving a curve? Well, recall that we can find that surface area is. By multiplying two pi by the definite integral between the limits of A and B of Y multiplied by the square root of 1 plus the derivative of y squared all with respect to X. Now do we have any of this information from our problem? Well, we are told that our lower bound A is equal to 2, our upper bound B is equal to 5, Y is equal to root X, which means that the derivative of Y. Is going to be equal to 1/2 of x raise to the part of negative. So if we substitute this into our formula for the surface area, that means S is going to be equal to 2 pi multiplied by the integral between 2 and 5 of root X multiplied by the square root of 1 + 1/2 of x to the negative half squared or with respect to X. Now before we go any further, let's try to simplify what's under our second expression with the root, OK? Now we know that it's 1 + 1/2 of X to the negative/2 squared. OK. So that would really be a 1/2 squared, which is a quarter, multiplied by X to the negative squared which is X-1, and we could rewrite that as 1 + 1 divided by 4 X. If we combine both terms into the same fraction, that is going to be equal to 4 X plus 1 divided by 4 X. But now remember all of this is under our root. Now why does, why is that important? Well, if I put back my root here, then notice that if we were to find, or if we were to split our route between our numerator and denominator, our numerator would be root 4 X plus 1, but our denominator would be the root of 4 X, which is equal to 2 root X. So if we take that, OK, and put it back into our surface area formula. Then S is going to be equal to 2 multiplied by a definite integral of root X multiplied by root 4 X + 1, OK, divided by 2 root X or with respect to X. Now this is pretty helpful because root X cancels root X and we can take our constant out of our integral to allow 2 to cancel 2 from the two pi and thus this is going to leave us with pi multiplied by our definite integral of route 4 X + 1 with respect to X, and this will be a bit easier to integrate. And because no, we can integrate this using substitution. So let's go ahead and do that. Let's let you be equal to 4 X plus 1. In that case, the derivative of you with respect to X will be equal to 4. Which means that DX will be equal to DU divided by 4. So now we have a substitute for both 4 X + 1 and DX in our expression for S. Next, remember we need to change the bones because originally these are the bones for our X terms. I know when X equals 2. You will be equal to 4 multiplied by 2 + 1, which is 9. While when X equals 5, U equals 4 multiplied by 5 + 1, which is 21. So in this case this means that our surface areas is going to be equal to pi multiplied by the definite integral between 9 and 21 of root U. Multiplied by DU divided by 4. Which we could rewrite as 1/4 of pi taking out our constant for or definite integral of you to the power of a half with respect to you. Now we can go ahead and integrate. So here we have a 4th of pi, and for our integral, the integral of you to the pore of 1/2 is equal to 2/3 of you to the pore of 3 halves, OK? And that's between our bones of 9 and 21. No, we can multiply our constant. We can take out or 2/3 here and multiply it by 14th of pi, which leaves us with a 16th of pi. And now if we substitute our bones, that will be 21 to the pore of 3 halves minus 9 to the pore of 3 halves inside our bracket. When we go ahead and simplify even further, OK, then we end up with our surface area to be a 6th of pi multiplied by 21 route 21 minus 27 square units. Thus, this is the surface area generated by revolving the curve Y equals root X above the X-axis between the bones of 2 and 5. Thanks a lot for watching everyone. I hope this video helped.

Find the area of the surface generated when the given curve is revolved about the given axis.

y=x^3/2−x^1/2 / 3, for 1≤x≤2; about the x-axis

Key Concepts

Surface Area of Revolution

Derivative of the Function

Integration over the Given Interval

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