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.

Graph showing region R bounded by y = 1/(1+x²), y = 0, x = 0, and x = 2, with a shell method illustration.

y = (1+x²)^−1,y = 0,x = 0, and x = 2; about the y-axis

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says let R be the region bounded by Y equal to 1 divided by the quantity of 2 + X2 in quantity, Y equal to 0, X equal to 0, and X equal to 1. Find the volume of the solid generated when R is revolved about the Y-axis using the shell method. And below the text, we are given a graph that has our function Y equal to 1 divided by the quantity of 2 plus X squad in quantity plotted on it. It also has X equal to 1 plotted on the graph. We also have our region R shaded on the graph, and R is bounded above by Y equal to 1 divided by the quantity of 2 plus X2 in quantity, bend below by Y equal to 0, is bounded on the left by X equal to 0, and bonded on the right by X equal to 1. We're also give them 4 possible choices as our answers. For choice A, we have 3 divided by 2, multiplied by pi, multiplied by the natural log of the quantity of 3 divided by 2 in quantity, cubic units. For choice B, we have 4 pi multiplied by the natural log of the quantity of 3 divided by 2 in quantity cubic units. For choice C, we have pi multiplied by the natural log of the quantity of 3 divided by 2 in quantity cubic units. And for choice D, we have 2 pi multiplied by the natural log of quantity of 3 divided by 2 in quantity cubic units. So, we're asked to find the volume that's generated by revolving the region R about the Y axis using the shell method. So, recall your formula for finding the volume using the shell method. Our volume V is going to be equal to. 2 i multiplied by the integral, and here we'll be looking at integrating over X, so our X values go from 0 to 1. And for the integran, we will have the radius of our shell. And we'll need to multiply this by the height of our shell. And since we're integrating X, we need to multiply this by DX. So we just need to determine the radius and height for each of our shells. So our radius. It's just gonna be the distance from the Y axis to our shell, so that's just gonna be equal to X since X is going to denote the position of our shell. And for the height of our shell, that's just going to be the distance between Y equals to 1 divided by the quantity of 2 + X2 in quantity and Y equal to 0. And so that difference is just 1 divided by the quantity of 2 + X2 in quantity. And so that means our volume V is going to be equal to. 2 i multiplied by the integral from 0 to 1. Of X multiplied by 1 divided by quantity of 2 plus X2 in quantity DX. So, now what we want to do is actually perform the integration, and we can use substitution to evaluate this integral. So, we're gonna set U equal to 2 + X2. So that means that DU is going to be equal to 2 XDX. So making these substitutions, we see that our volume V is going to be equal 2, and we're going to pull the factor of 2 inside the integral to combine it with the X and the DX to give us our DU. So out in front, the only factor that's going to be left is just going to be pi, and this is going to be multiplied by the integral, and here we have to change our limits of integrations we're changing variables. So when X is equal to 0 for the lower limit, we see that U is going to be equal to 2 + 0 squad, which is going to be 2. And when X is equal to 1 for the upper limit, we see that U is going to be equal to 2 + 1 squared, which is 3. And then for the integram, we will have 1 divided by U multiplied by D U. And so now we just need to perform the integration, so this is going to be equal to high. Multiplied by, and here we are integrating one divided by you with respect to you, and recall that that's going to give us the natural log of the absolute value of you, and this needs to be evaluated from 2 to 3. So substituting in the limits of integration, we see that our volume is going to be equal to high, multiplied by the quantity of the natural log of 3 minus the natural log of 2 in quantity. And here we've dropped the absolute value sign, since both 3 and 2 are positive quantities. And we can combine our two natural log terms using the properties of logarithms. In doing so, we see that our volume B is going to be equal to pi, multiplied by the natural log of the quantity of 3 divided by 2 in quantity. Cubic units. 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 C. 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.

y = (1+x²)^−1,y = 0,x = 0, and x = 2; about the y-axis

Key Concepts

Shell Method

Volume of Revolution

Definite Integrals

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