87. Surface area Find the area of the surface generated when t...

Question

87. Surface area Find the area of the surface generated when the curve f(x) = sin x on [0, π/2] is revolved about the x-axis.

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says what is the surface area of the solid form by revolving F of X, which is equal to cosine of x, on the closed interval from 0 to pi divided by 3 on the X-axis. So first step is to recall a formula for finding the surface area of a solid formed by revolving a function about the x-axis. And to recall that our surface areas is going to be equal to 2 pi multiplied by the interval from A to B, of F of X. Multiplied by the square root of the quantity of 1 plus the quantity of F of X. In quantity squared in quantity DX. So, in our case, we're looking at the interval from 0 to pi divided by 3, so that means that A is going to be equal to 0, B is going to be equal to pi divided by 3. And we're given the function of F of X in the problem. So F of X is going to be equal to cosine of X. Now we just need to take a derivative of this function in order to find F. So taking a derivative of F with respect to X. And so we'll have F of X is going to be equal to minus sine of X. So substituting in these values into our Formula 4 S, we see that S is going to be equal to 2 pi. Multiplied by the integral from 0 to pi divided by 3 of cosine of X multiplied by the square root of the quantity of 1 plus squad of X in quantity DX. So now we just have to evaluate this integral, and we're going to begin by evaluating it using substitution. So we're going to set U equal to sin of x, and that means that DU is going to be equal to cosine of X. DX. So making this substitution, we see that S is going to be equal to 2 pi multiplied by the integral, and here we have to change our limits of integration. So when X is equal to 0, U is going to be equal to the sine of 0, which is going to be 0. And when X is equal to pi divided by 3, we'll have that U is equal to the sign of pi divided by 3, which is the square root of 3 divided by 2. And then for the integran, once we do the substitution, we'll just have the square root with the quantity of 1, plus U squared. In quantity to you. And we can evaluate this integral using one of our standard formulas for integration. So we call that the interval of the square root of the quantity of a quad plus Y2 in quantity D Y is equal toy divided by 2 multiplied by the square root of the quantity of a2 plus Y2 in quantity. Plus, a 2 divided by 2. Multiplied by the natural log of the absolute value of the quantity of the square root of the quantity of A2d plus Y2 in quantity. Plus Y in quantity plus D. So that means that S is going to be equal to 2 pi multiplied by the quantity of U divided by 2, multiplied by the square root of the quantity of 1 plus U squared in quantity. Plus, 1 divided by 2 multiplied by the natural log of the absolute value of the square root of the quantity of 1 plus U squared in quantity, plus U in quantity. And all that is evaluated on the interval from 0 to the square root of 3 divided by 2. So, in our expression here, we can factor out a 1/2 out of each of the terms of the brackets, and that will cancel out with a 2 in the front. And so that means that this is going to be equal to. Pi multiplied by the quantity, and here we'll evaluate our expression at our two boundaries. So for the upper boundary, we'll just have the square root of 3 divided by 2 multiplied by the square root of the quantity of 1 plus the quantity of the square root of 3 divided by 2 in quantity squared in quantity. Plus, the natural log of the absolute value of the square root of the quantity of 1, plus the quantity of the square root of 3 divided by 2 in quantity, squared in quantity. Plus, the square root of 3, divided by 2, in quantity. Then we'll have minus for our lower bound, the quantity of 0, multiplied by the square root of the quantity of 1 + 0 squared. In quantity Plus The natural law of the absolute value of the square root of the quantity of 1 plus 0 squared in quantity plus 0 in quantity. And so now we just need to simplify this expression. So doing so, this is going to be equal to high, multiplied by the quantity of the square root of 21, divided by 4. Plus The natural log of the absolute value of the square root of 7 divided by 2. Plus, the square root of 3, divided by 2 in quantity, then we'll have minus. 0 minus the natural log of 1. In quantity So, to remove the fractions from the terms inside the brackets, we'll factor out a 1 divided by 4 out of each of those terms. In doing so, this is going to be equal to pi divided by 4, multiplied by the quantity of the square root of 21. Plus 4 multiplied by the natural law of the quantity, and here we can remove the absolute value sign, since everything inside there is a positive quantity. So we'll have the square root of 7, plus the square root of 3, and that quantity is going to be divided by 2 in quantity. And the remaining terms are are 0, so we don't need to include them since the natural log of 1 is 0. So this here is our final answer that we have pi divided by 4, multiplied by the quantity of the square root of 21, plus 4 multiplied by the natural log of the quantity of the square root of 7, plus the square root of 3 divided by 2 in quantity. So this is the answer that we're looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

87. Surface area Find the area of the surface generated when the curve f(x) = sin x on [0, π/2] is revolved about the x-axis.

Key Concepts

Surface Area of Revolution

Definite Integral

Derivative and its Role in Surface Area

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