Compute the surface area of revolution about the x -axis over the interval. y=sin x, [0,

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

Key Concepts

Surface Area of Revolution

This concept involves calculating the area of a surface obtained by revolving a curve around an axis. The computation uses the formula S = 2??[a to b] f(x)?(1 + [f?'(x)]²) dx when revolving around the x-axis, where f(x) represents the function and f?'(x) its derivative. It is an integral application illustrating how integration can be used to measure complex shapes.

Definite Integration

Definite integration is used to evaluate the accumulated sum, such as area or volume, over a specific interval [a, b]. In the context of surface area of revolution, the definite integral sums the contributions of each infinitesimal element of the surface over the entire interval, providing a precise measure of the total area.

Differentiation

Differentiation is used to determine the rate of change of a function. When computing the surface area of revolution, the derivative f?'(x) is critical as it is used inside the square root in the integrand to account for the slope of the curve, which influences the length of the surface element being revolved.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, often appear in calculus problems involving curves and their derivatives. Their properties and relationships are essential in setting up and simplifying the expressions within the integral for problems related to areas, lengths, or volumes of shapes generated by revolving trigonometric curves.

Transcript

00:01 In this problem, we are going to compute the surface area of sine of x from 0 to pi when it's revolved about the x -axis.

00:08 And when we revolve this function about the x -axis, we get a surface that looks somewhat like this to the right of our graph.

00:19 Well, we know that surface area is the integral, in this case from 0 to pi of 2 pi, which is sine of x.

00:38 Times the square root of 1 plus y prime squared.

00:43 So we get 1 plus cosine squared x dx.

00:50 And from here we can let u be cosine of x.

00:59 So minus du is sine x d x.

01:11 And now we can pull out sine x d x and replace that by minus to u.

01:16 And we can also change our limits of integration.

01:21 So this becomes the integral from actually from 1 to minus 1 of minus 2 pi times the square root of 1 plus u squared d u and if we flip bar limits of integration we can get rid of this minus sign and then from here we can take advantage of the fact that 1 plus tangent squared is secan squared so we can let u be tangent theta so d u is secan squared so d u is d theta.

02:32 And we can also change our limits of integration again.

02:40 So this is the integral from minus pi on four, pi on four of two pi times the square root of 1 plus tangent square theta.

02:57 Well again one plus tangent square theta is secant square theta and if we take the square root of that we get secant theta, but then we have to multiply by secant squared theta d theta.

03:20 So altogether this is 2 pi times secant cube theta d theta.

03:28 And then the integral of secant cube theta d theta, we can actually use the reduction formula or even integration by parts, but we get that this is one half times secant theta, tangent theta plus one half times the natural log of seekin theta plus tangent theta plus c.

04:20 But of course in this case we have limits of integration.

04:25 This is a definite integral...

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