Compute the surface area of revolution about the x -axis over the interval. y=e^-x, [0,1]

Name: Problem 38, Chapter 8: Calculus for AP
Uploaded: 2020-09-27T17:03:36-08:00
Duration: 11 min 42 s
Channel: Melvin Adkins
Description: Compute the surface area of revolution about the x -axis over the interval. y=e^-x, [0,1]

Compute the surface area of revolution about the x -axis...

Key Concepts

Definite Integration

Definite integration is used to sum the contributions of infinitely many small elements over a specific interval. In the context of computing a surface area of revolution, it helps accumulate the results from the differential surface areas along the curve between given bounds, providing the total surface area generated by the rotation.

Differentiation of Functions

Calculating the derivative of the function is a critical step because it provides the necessary information for the arc length differential. The derivative measures the rate of change of the function with respect to its variable, and when squared and added to one, it forms an integral part of the formula that ‘stretches’ the contribution of each infinitesimal segment in the rotation.

Arc Length Differential

The term ?[1 + (dy/dx)^2] represents the differential element of arc length along the curve. It quantifies how much the curve deviates from a straight line by incorporating the slope at each point, thereby adjusting the integration to accurately account for the actual path length as opposed to mere horizontal displacement.

Surface Area of Revolution

This concept involves finding the area of a surface generated when a curve is revolved around an axis. It requires setting up an integral using the formula S = 2??(function value)?[1 + (derivative of function)^2] dx (or the corresponding version for y and x). The integral sums up the contributions of the rotated infinitesimal elements along the curve.

Transcript

00:01 In this problem, we will compute the surface area of e to the minus x from 0 to 1 when it's revolved about the x -axis.

00:10 And when we revolve this function about the x -axis, we get the figure to the right.

00:18 And we know that the surface area is equal to the integral from 0 to 1 of 2 pi -y, which is e to the minus x, times the square root, of 1 plus y prime squared.

00:39 Y prime is minus e to the minus x.

00:42 So y prime squared is just e to the minus 2x.

00:50 And then we also have dx.

00:54 So now that we have that, we can use a u substitution.

00:59 So we can let you be e to the minus x, which means that minus d u is e to the minus x.

01:12 Dx and we can also change our limits of integration and also e to the minus x is minus e to the minus x times d x is minus d u so we get that this is the integral from one to one over e of minus two pi times the square root of one plus u squared d u and we can get rid of this minus sign by flipping our limits of integration.

02:19 And then from this step, we can also do a trick substitution by allowing you to be tangent theta, which means that du is sequence squared theta, d theta.

02:49 And again, we can change our limits of integration.

02:55 So we get that this is the integral from the inverse tangent of 1 over e to pi on 4 of 2 pi times the square root of 1 plus tangent squared theta times secan squared theta d theta and 1 plus tangent squared theta is secant squared theta so we get secant squared theta we get the square root of secan square theta which is just secant theta which means that this is the integral of secund cubed theta d theta and then to solve for this integral we could use integration by parts or we could use the reduction formula for secant theta but when we solve this integral we get that this is we also have this two pi that we don't want to forget so let's write that in there but we can factor out this two pie and we get that this is 1 1�t times secant of pi on 4 times tangent of pi on 4 plus 1 half times the natural log of secant of piom 4 plus tangent of pi on 4 minus the quantity 1 half times secant of inverse tangent of 1 over e times tangent of inverse tangent of 1 over e plus 1 half times the natural log of secant of inverse tangent of 1 over e plus tangent of inverse tangent of 1 over e right so this is 2 pi let's see secant is 1 over cosine so sikin of pi on 4 that is just root 2, tangent of pi on 4 is 1, and this is plus 1 half times the natural log of root 2 plus 1, minus 1 half.

07:24 So let's evaluate this secant of inverse tangent of 1 over e, and let's do that up here...

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