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=1/4(e^2x+e^−2x), for −2≤x≤2; about the x-axis

Accepted Answer

Welcome back everyone. In this problem, we want to determine the surface area formed by revolving the curve Y equals 1/8 of E 4 X plus E 4 X for X greater than or equal to -4, but less than or equal to 4 above the X axis. Now what do we know about finding the surface area of a curve revolving above the X axis? Well, recall That the surface areas is going to be equal to 2 pi multiplied by the definite integral between the limits A and B of Y multiplied by the square root. Of 1 plus the derivative of y with respect to x. All squared, all with respect to X. Now what do we know that can help us to solve? Well, so far we know, OK, that a. Or lower limit equals -4. B or upper limit equals 4. Y or function is an e e 4 X plus e negative 4 X, which means that our derivative of Y with respect to X is going to be equal to an if of 4 E 4 X minus 4 E 4 X. And then we simplify that, it equals a half of E4 X. Minus e to the negative 4X. So now we have all we need to figure out our surface area. So let's plug these values into our formula. No, this tells us that the surface area equals 2 pi multiplied by the integral between -4 and 4. Of 1/8 of E to the forex plus E to the negative 4X, OK. Multiplied by the square root of one plus. A half of e to the 4 X minus e to the negative 4 X all squared, all with respect to x. So where do we go from here? Well, first, let's start by trying to simplify our root, OK? That might make it easier for us to solve. So now we are simplifying the root. Now we know that under the root, we have the expression 1 plus the derivative of Y with respect to X squared. OK. So, no, really, when we square a half of E 4 X minus E to the negative 4 X, then we're going to end up with a quarter of E to the power of 8 X minus 2 plus E to the power of negative 8X. And if we combine both of these terms, in other words, we factor to a quarter, then we're going to have a quarter of E to the 8 X plus e to the negative 8 X minus 2 + 4, which we know minus 2 + 4 E equals 2. So we have all of this being multiplied by 25%. So now if we put this into a root, this would imply then that the root of 1 plus the derivative of y with respect to X squared. is going to be equal to the square root, OK, of our expression, a quarter of E to the 8 X plus E to the negative 8X. OK. Or say Yes, E to the 8 X, my apologies, plus E to the negative 8X + 2. Well now, there's something else that can help us. We also know an identity for our expression here. Recall that this is a known identity, OK? So we know that E 8 X + E to the negative 8X + 2 is actually the perfect square for E 4X plus E to the negative 4X. OK? I know if we substitute that into our expression. Then this is going to be a quarter of, sorry, the square root. Of a quarter of year. To the 4 X plus e to the negative 4 X squared and now we can apply our root to both terms to leave us with a half of e to the 4 X plus e to the negative 4 X. So our square root for all of this term eventually works out to this final expression that we have. And now we can put this back into our formula for the surface area because no, this means the surface area will be equal to 2. Multiplied by the definite integral between -4 and 4. Of an 8 of the 4 X plus E to the negative 4X. OK. Multiplied by a half of E 4 X plus e-4 X. I with respect the X. If we do some simplifying here, this works ought to be 2 multiplied by a definite integral from -4 to 4 of an 8th multiplied by 1/2, that's 1/16 of E 4 X plus e to the negative 4 X squared with respect to X. Now if we expand our squared term, OK, so no and multiply 2 by 2 by 1/16, we're going to have an 18 of pi multiplied by our definite integral of e 8 X plus 2 plus e to the negative 8X all with respect to X. Now where do we go from here? Well, at this point we can evaluate each integral to help us solve. So now by evaluating each integral, OK. Then for our first term, and that is the definite integral we're finding for each. For our first term, OK, we're gonna have between -4 and 4. Of e to the 8 X with respect to X. And that is gonna be equal to 1/8 of E to the 8 X. OK. Between 4 and -4, and when we solve, that's going to be equal to. 1/8 of E to the power of 32 minus E to the power of -32. Next, for our constant term 2, OK, we will find a definite integral of that with respect to X, it's going to be equal to 2 multiplied by 4 minus -4. That's 2 multiplied by 8, which equals 16. Then finally, for our definite integral or e to the negative 8 X with respect to x. That's going to be equal to a negative/8 of e-8X between -4 and 4, which would also be equal to 1/8 of E to the pore of 32 minus e to the negative pore of 32. OK, so no or definite integral will be the sum of all these three terms. So that means our surface area will be equal to 1/8 of pi multiplied. By 1/8 of E to the 32 minus E to the negative 32 twice, so that's 2 multiplied by 1/8 of E to the 32 minus E to the negative 32. OK. Plus. 16 And if we distribute 1/8 of pi across both terms for our first term, 1/8 of pi multiplied by 2 equals 4th of pi multiplied by 8 equals 132 or 1/32 of pi or pi divided by 32. So that's pi divided by 32 multiplied by e to the pore of 32 minus e to the negative 32 plus an 1/8 of pi multiplied by 16, which is 2 pi. Thus, the surface area is equal to pi divided by 32 multiplied by E 32 minus E 32 plus 2 square units. 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=1/4(e^2x+e^−2x), for −2≤x≤2; about the x-axis

Key Concepts

Surface Area of Revolution

Parametric Derivatives

Definite Integrals

Watch next