9-34. Shell method Let R be the region bounded by the followin...

Question

9-34. Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about indicated axis.

{Use of Tech} y = √50 -2x², in the first quadrant; about the x-axis

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says let R be the region bounded by the curve Y is equal to the square root of the quantity of 75 minus 3 X squad in quantity in the first quadrant. Find the volume of the solid generated when R is revolved about the X-axis using the shell method. And below the text, we were given a graph, and on this graph we have our curve Y is equal to the square root of the quantity of 75 minus 3 X 2d and quantity plotted on it. We also have the region R shaded on this graph, and that region is bounded by our curve Y, as well as the x-axis and the Y axis. So we're asked to find the volume of the solids generated when R is revolved about the X-axis using the shell method. Now since we're using the shell method and revolving about the X-axis, we're going to be integrating with respect to Y. So, in order to find our limits of integration, we'll first need to find our Y intercept. So that means that we need to set X equal to 0 and solve for Y for a curve Y. In doing so, we will have Y as equal to the square root of the quantity of 75 minus 3 multiplied by 0 squared in quantity. And if I wait this expression, we see that y is going to be equal to 5 multiplied by the square root of 3. Now, since we are also integrating with respect to Y, we need to have all of our functions as functions of Y. So we need to take our curve and solve it for X so that we'll have X as a function of Y. So doing so, we'll start off with our equation for a curve, which is Y is equal to the square root of the quantity of 75 minus 3 x 2 and quantity and solve this for X. So squaring both sides will get Y2 is equal to 75 minus 3 x 2. Now adding 3x2 to both sides of the equation as well as subtracting Y2d from both sides, we will get 3 x 2 is equal to 75 minus Y2. Dividing both sides of the equation by 3, we'll get that x 2 is equal to the quantity of 75 minus y2d and quantity divided by 3. Then taking the square root of both sides of this equation, we'll get that X is equal to the square root of the quantity of the quantity of 75 minus y2d and quantity divided by 3 in quantity. And here we've just taken the positive square root. Since we're in the first quadrant. Next, we need to determine the radius and height of each of our shells. So, we'll begin by looking at the radius of each shell. And the radius here is just our distance from the X axis to the shell. And so this is just going to be equal to our coordinate. Y. Now, for the height of our shell, that's just gonna be equal to the X value that bounds our region R on the right, minus the X value that bounds our region R on the left. So, on the right, Our region R is bounded by X, which is equal to the square root of the quantity of the quantity of 75 minus Y2 in quantity divided by 3 in quantity. And then we'll have minus n on the left. Our region R is bounded by X equal to 0, so we'll have minus 0. So that means our height is just going to be equal to the square root of the quantity of the quantity of 75 minus Y2 in quantity divided by 3 in quantity. So now we have all the information that we need to set up our integral for our volume. So, we call that the volume, using the shell method, that our volume V is going to be equal to 2 pi, multiplied by the integral, and here we're integrating with respect to Y. So Y is going to go from 0 to 5 multiplied by the square root of 3. And then for integram, we will have our radius, which is Y, multiplied by our height, which is the square root of the quantity of 75 minus Y2 in quantity divided by 3 in quantity, and that gets multiplied by Dy since we're we're integrating with respect to Y. So now we just need to evaluate this integral, and we'll evaluate this using substitution. So we'll set U equal to 75 minus Y2d, and that means that DU is going to be equal to minus 2 multiplied by Y multiplied by DY. Now, if we look at our integrated here, we actually just have 2 multiplied by Y multiplied by DY. So that means that minus DU is going to be equal to 2 YDY. And so now we can make this substitution into our integral. In doing so, we see that V is going to be equal to, and we'll actually factor out a 1 divided by the square root 3 out of our integram and pull it out in front of the integral. So we will have minus pi divided by the square root of 3 as coefficient out front, where the minus sign is coming from the minus DU that we'll be using later. And that gets multiplied by the interval, and here we need to change our limits of integration, since we're changing variables. So for the lower limit of integration, when Y is equal to 0, U is equal to 75 minus 0 squad, which is 75. And then for our upper upper limit of integration, when Y is equal to 5 multiplied by the square root 3, U is going to be equal to 75 minus the quantity of 5 multiplied by the square root of 3 in quantity squared, which gives us 0. Then for the integrand, we'll just have the square root of U D U. So now, we just need to do the integration with respect to you. And the first thing we want to do is to switch our limits of integration. And recall that when we switch the limits of integration that we need to multiply our integral by -1. And so in doing so, this is going to be equal to pi divided by the square root of 3, multiplied by the integral going from 0 to 75 of the square root of u to U. So now performing the integration, we see that this is going to be equal to pi divided by the square root of 3. Multiplied by, and here we're going to be integrating the square view with respect to you and recall that when we do that integration, we get that you, we have you raised to the quantity of 3 divided by 2 in quantity, divided by the quantity of 3 divided by 2 in quantity, and this needs to be evaluated from 0 to 75. So substituting in the limits of integration, we see our volume V is going to be equal to, and here we'll simplify the coefficient out in front, and we will have 2 pi divided by the quantity of 3 multiplied by the square root of 3 in quantity, and this gets multiplied by the quantity of 75 raised to the 3 divided by 2. And then we'll have minus 0 raised to the quantity of 3 divided by 2 in quantity. So simplifying this expression, this is going to be equal to 2 pi divided by the quantity of 3 multiplied by the square root of 3 in quantity, multiplied by. 375 multiplied by the square root of 3. And finally, evaluating this compression, we see that our volume V is going to be equal to 250 multiplied by pi cubic units. So this here 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.

9-34. Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about indicated axis.

{Use of Tech} y = √50 -2x², in the first quadrant; about the x-axis

Key Concepts

Shell Method for Volume

Region Bounded by Curves

Revolution About the x-axis

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