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.

x=√12y−y^2, for 2≤y≤10; about the y-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 X, which is equal to the square root of the quantity of 8 Y minus Y2 in quantity for 1 less than or equal toy, which is less than or equal to 7, is revolved about the y axis. And we're given 4 possible choices as our answers. For choice A, we have 8 pi square units. For choice B, we have 48 pi square units. For choice C, we have 84 pi square units, and for choice D, we have 44 pi square units. So we need to determine the area of a surface generated when a curve is revolved about the y axis. So here we need to recall our formula for finding that surface area. So we'll call it surface areas. We recall that this is going to be equal to 2 pi multiplied by the interval of X, multiplied by the square root of the quantity of 1 plus the quantity of the derivative of X with respect to Y in quantity squared in quantity D Y. So the first thing we need to do is find the derivative of X with respect to Y. And this is going to be equal to the derivative with respect toy of our function X, which is the square root of the quantity of 8 Y minus Y squad in quantity. So taking this derivative, this is going to be equal to. And here we call that when you take the derivative of the square root function, you'll get 1 divided by the quantity of 2 multiplied by the square root of the quantity of 8 Y minus Y squared in quantity, and this gets multiplied by the derivative of the argument. Of our square root functions. So, Taking the derivative with respect toy of the quantity of 8 Y minus Y2d, we get the quantity of 8 minus to Y in quantity. So here we just had to apply our chain rule. And we can actually simplify the expression. So in the numerator we can factor out a 2 out of our two terms of the numerator. With that factor of 2 that we have factored out, we'll cancel out with a factor of 2 in the denominator. So that means that the derivative of X with respect to Y is going to be equal to the quantity of 4 minus Y in quantity, divided by the square root of the quantity of 8 Y minus Y squared in quantity. However, in our formula for our surface area, we actually want the square of this derivative. So the quantity of the derivative of X with respect to Y in quantity squared is going to be equal to the quantity of 4 minus Y in quantity squared divided by the quantity of 8 Y minus Y2. In quantity. So substituting this result back into our surface area formula, we see that S is going to be equal to 2 pi multiplied by the interval from 1 to 7, since Y takes on the values between 1 and 7. And then inside the integral we'll have X, which is the square root. Of the quantity of 8 Y minus Y2 in quantity that gets multiplied by the square root of the quantity of 1 plus the quantity of 4 minus Y in quantity squared divided by the quantity of 8 Y minus Y2 in quantity, and all of that is going to be integrated with respect to Y. So now we can actually combine the two square roots in our integran. In doing so, we see that S is going to be equal to 2 pi, multiplied by the integral from 1 to 7. Of the square root. Of the quantity of 8 Y minus y2 plus the quantity of 4 minus Y in quantity squared in quantity. So now we just need to multiply out the binomial inside the square root. In doing so, we'll see that S is equal to 2 pi multiplied by the integral from 1 to 7 of the square root of the quantity of 8 Y minus Y2 + 16. Minus 8 Y plus y squad in quantity D Y. So, simplifying the polynomial in the underneath the radical, we see that S is going to be equal to 2 pi, multiplied by the integral from 1 to 7 of the square root of 16 DY. So now we just need to perform the integration, and we can actually factor out the square root of 16, which is equal to 4, out in front of the integral since it's a constant, and so we'll have 4 multiplied by 2 pi. So that is going to be equal to 8 pi and here we're just going to be integrating 1 with respect to Y, which will give us Y, and this needs to be evaluated from 1 to 7. And so, substituting in our limits of integration, we see that this is going to be equal to 8 pi, multiplied by the quantity of 7 minus 1 in quantity, and simplifying this answer, we see that our surface area is going to be equal to. 48 pie square units and so that is the answer that we were looking for for this problem. And if we compare our answer here to our answer choices, we see that the correct answer choice corresponds to answer choice B. 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.x=√12y−y^2, for 2≤y≤10; about the y-axis

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

x=√12y−y^2, for 2≤y≤10; about the y-axis