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=8√x, for 9≤x≤20; about the x-axis

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says determine the area of the surface generated when the curve Y is equal to 10 multiplied by the square root of X. 41 is less than equal to X, which is less than equal to 5, is revolved about the X axis. So to begin, we need to recall our formula for finding the area of surface generated when a curve is revolved about the X axis. And so that surface area, which we'll denote as S, is going to be equal to. To pie Multiplied by the integral of Y, multiplied by the square root of the quantity of 1 plus the quantity of the derivative of Y with respect to X in quantity squared in quantity D X. So in order to use this formula, we'll need to first find the derivative of Y with respect to X. So the derivative of Y with respect to X is going to be equal to the derivative with respect to X. Of the quantity of 10 multiplied by the square root of X. And this is going to be equal to 10 is constant, so it can move through the derivatives, so we'll have 10, and here we're gonna be multiplying this by the derivative with respect to X of the square root of X, and recall for that derivative, we get 1 divided by the quantity of 2 multiplied by the square root of X in quantity. And just simplifying this expression, we see that this is going to be equal to. 5 divided by the square root of X. And we actually want the quantity of the derivative of Y with respect to X in quantity squared, and so that's just going to be equal to the quantity of 5 divided by the square root of X in quantity squared, which is going to be equal to 25 divided by X. So that means that our surface area is going to be equal to 2 pi, multiplied by the integral, and here for Y, we will have 10 multiplied by the square root of X in quantity, and that gets multiplied by the square root of the quantity of 1 plus 25 divided by X in quantity d X, and the limits of integration are going to go from 1 to 5, since X is supposed to be between 1 and 5. So now we just need to simplify this expression. So this is going to be equal to. We have 10 as a constant inside the integral, so we'll factor it out and multiply it by the 2 pi that's already in front of the integral. And so we'll have 20 pi, multiplied by the integral from 1 to 5. Of and here we'll bring the square root of X inside the other square root term and in doing so it'll just become x underneath the radical. So we'll have the square root of the quantity of X plus 25 in quantity d X. So now we just need to evaluate this integral, and we can evaluate it using U substitution. So we'll set u equal to x plus 25, and that means that DU is going to be equal to DX. So, our integral becomes equal to here we still have the same constant out front, so we'll have the 20 pi that gets multiplied by the integral, and here, since we're changing variables, we need to change the limits of integration. So when X is 1, U is going to be equal to 1 + 25, which is going to be 26. And when X is equal to 5, U is going to be equal to 5 plus 25, which is 30. And then we'll have the square root of U D U. So now we just need to perform the integration. So this is going to be equal to 20 pi. Multiplied by the quantity, and here when we integrate the square view, we will have you to the 3 house power divided by the quantity of 3 divided by 2 in quantity, and this needs to be evaluated from 26 to 30. So now we just need to substitute in our limits of integration. So that means our surface area is going to be equal to. And here we're going to simplify the fraction of 1 divided by the quantity of 3 divided by 2 in quantity, and multiply that by the 20 pi. In doing so, we will get 40 pi divided by 3. Multiplied by the quantity, and here we'll substitute in our limits and integration. So, for the upper limit of integration, we will have 30, and that gets raised to the 3 has power, then we'll have minus for the lower limit of integration, we'll have 26 raised to the 3 halves power. In quantity, and we can simplify this expression a little bit. So this is going to be equal to 40 pi divided by 3, multiplied by the quantity, and 30 raised to the 3 has power is going to be 30, multiplied by the square root of 30. Then we'll have minus and 26 raised to the 3 has power, we can rewrite as 26 multiplied by the square root of 26 in quantity. Now, for the quantity inside the princes, we can factor out a 2 from each of those terms and pull it outside. That the prince sees in doing so, we'll see that our final answer for our surface area is going to be equal to. 80 pie Divided by 3, multiplied by the quantity of 15, multiplied by the square root of 30, minus 13, multiplied by the square root of 26, in 20 squared units. And so that is the answer that we were looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Find the area of the surface generated when the given curve is revolved about the given axis.y=8√x, for 9≤x≤20; about the x-axis

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

y=8√x, for 9≤x≤20; about the x-axis