A 1.5-mm layer of paint is applied to one side of the followin...

Question

A 1.5-mm layer of paint is applied to one side of the following surfaces. Find the approximate volume of paint needed. Assume x and y are measured in meters.

The spherical zone generated when the curve y=√8x−x^2 on the interval 1≤x≤7 is revolved about the x-axis

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says a 1 millimeter thick film is applied to one side of the surface generated by revolving the curve Y, which is equal to the square root of the quantity of 12 X minus X squad in quantity, or 3 is less than equal to X, which is less than equal to 9, about the x-axis. What is the approximate volume of film required? All measurements are in meters. So for this problem, we need to find the volume of the film, and the way we're gonna find the volume is, first, we're gonna find the area of the surface that's generated by revolving our curve about the X-axis. Then we'll take that surface area and multiply it by the thickness in order to get our volume. So, we first need to find the surface area. So recall your formula or finding the area of the surface that's generated by revolving the curve about the X-axis. So that surface area S is going to be equal to 2 pi, multiplied by the integral, and here our limits of integration are going to go from 3 to 9, since our values of X. Are between 3 and 9 given in the problem. Then for the integrand, we will have 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 DX. So our next step is to calculate the derivative of Y with respect to X. And this is going to be equal to the derivative with respect to X of Y, which is the square root of the quantity of 12 x minus X squad in quantity. So taking this derivative, this is going to be equal to. He recall that when you take the derivative of a square root function, you get 1 divided by the quantity of 2 multiplied by the square root functions will have the square root of the quantity of 12 X. Minus x 2 in quantity, but we need to apply our chain rule, so we need to multiply this by the derivative with respect to x of the quantity of 12 x minus X squared in quantity, and taking that derivative, we'll be multiplying this fraction by the quantity of 12 minus to x in quantity. And we can simplify this expression in doing so, this is going to be equal to the quantity of 6 minus X in quantity divided by the square root of the quantity of 12 X. Minus x 2 in quantity. So substitute that result back into our expression for S, we see our surface area is going to be equal to 2 pi multiplied by the integral from 3 to 9 of Y, which is the square root of the quantity of 12 X minus X squad in quantity, that gets multiplied by the square root of the quantity of 1. Plus, and here we need to square our result for the derivative of Y with respect to X. In doing so, we'll have the quantity of 6. Minus X in quantity squared, divided by the quantity of 12 X minus X2, in quantity DX. So simplifying our integrated by multiplying together our two square roots. Doing so, we'll have that S is going to be equal to 2 pi, multiplied by the integral from 3 to 9 of the square root. Of the quantity of 12 X minus X squared. Plus, the quantity of 6 minus X in quantity squared in quantity DX. Now we need to multiply out the binomial inside the square root sign. In doing so, our surface area is going to be equal to 2 pi, multiplied by the integral from 3 to 9 of the square root. Of the quantity of 12 X minus X2. Plus 36 minus 12 X. Plus x 2 in quantity. DX. And finally simplifying our expression, we see that this is going to be equal to 2 pi, multiplied by the interval from 3 to 9 of the square root of 36. DX. Now, the square root of 36 is just going to be equal to 6, which is a constant, and we can pull it out front of the integral. In doing so, the coefficient out front is going to become 12 multiplied by pi, and here we're gonna be multiplying the integral from 3 to 9 of DX. And so we integrate essentially one with respect to X, we get X. And so we'll multiply our 12 pi by X and we'll need to evaluate that from 3 to 9. So substituting in our limits of integration, this is going to be equal to. 12 pi multiplied by the quantity of 9 minus 3. In quantity, and evaluating this question, we see that our surface area is going to be equal to. 72 pi and this has units of meters squared, since we're told that all of our units are measured in meters. So now we need to use the surface area to calculate our volume. Though our volume, V, is going to be equal to our surface areas multiplied by our thickness which will denote as T. Now we're told the problem that, our thickness T is equal to 1 millimeter. But we'll need this in meters, so we'll divide this by 1000, and so that means that our thickness T is going to be equal to 0.001 m. So for volume, we will have V is going to be equal to. 72 pie. Meters squared. And that gets multiplied by 0.001 m. And so evaluating this expression, we see that our volume V is going to be equal to 0.23 m cubed. So here we've just round it to two decimal places, and this is the answer that we are looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

A 1.5-mm layer of paint is applied to one side of the following surfaces. Find the approximate volume of paint needed. Assume x and y are measured in meters. The spherical zone generated when the curve y=√8x−x^2 on the interval 1≤x≤7 is revolved about the x-axis

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A 1.5-mm layer of paint is applied to one side of the following surfaces. Find the approximate volume of paint needed. Assume x and y are measured in meters.

The spherical zone generated when the curve y=√8x−x^2 on the interval 1≤x≤7 is revolved about the x-axis