120. Equal volumes
a. Let R be the region bounded by the...

Question

120. Equal volumes

a. Let R be the region bounded by the graph of f(x) = x^(-p) and the x-axis, for x ≥ 1. Let V₁ and V₂ be the volumes of the solids generated when R is revolved about the x-axis and the y-axis, respectively, if they exist. For what values of p (if any) is V₁ = V₂?

b. Repeat part (a) on the interval [0, 1].

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says consider the region R bounded by the curve Y is equal to E raised to the quantity of minus KX in quantity and the X-axis, or X greater than equal to 0, where K is greater than 0. Let V1 and V2 denote the volumes of the solids obtained by revolving R about the X-axis and the Y axis respectively. For which value of K, if any, do we have V1 is equal to V2? If we give 4 possible choices as our answers. For choice A, we have 16, for choice B, we have 25. For choice C, we have 4, and for choice D, we have no such value if K exists. So we're gonna begin just by calculating the two volumes, and we're going to begin with V1. And so V1 here is the volume of the solid obtained by revolving R about the X-axis. So we're going to use the washer method to find this volume. So we call for the washer method that V1 is going to be equal to pi. Multiplied by the integral from 0 to infinity, since X is just greater than or equal to 0. Multiplied by the quantity, and here we have to take the difference of the two radii squared. So, for our out radius, that's just gonna be our curve Y. Which is E rates to the quantity of minus K multiplied by X in quantity. So we'll need to square that, and so we'll have E rates to the quantity of minus 2 K multiplied by X in quantity. Then we'll have minus for inner radius. That's just going to be our X-axis, which is at Y equal to 0, so we'll have 0 squared for our inner radius in quantity, and that gets multiplied by DX. So now we need to simplify this expression. The V1 is going to be equal to, and we noticed that we have an improper integral. So we're going to rewrite this as the limit. As R approaches infinity. Of pi multiplied by the integral from 0 to R. Of E raised to the quantity of minus 2 K multiplied by X and quantity DX. So now we just need to perform the integration. So this is going to be equal to the limit, as R approaches infinity of pi. Multiplied by And here, when we integrate E raised to the quantity of minus 2k multiplied by X in quantity, we get the quantity of minus 1 divided by the quantity of 2k in quantity multiplied by E rates to the quantity of minus 2k multiplied by X in quantity, and this needs to be evaluated from 0 to R. So substituting in the limits of integration, we see that this is going to be equal to the limit. As are approaches infinity. Of pie Multiplied by the quantity of minus 1 divided by the quantity of 2k in quantity, multiplied by Erates to the quantity of minus 2 K multiplied by R in quantity. Then we'll have plus 1 divided by the quantity of 2K in quantity multiplied by E, raised the quantity of minus 2K multiplied by zero in quantity. Now, as R approaches infinity, E reads to the quantity of minus 2K multiplied by R also approaches 0. So, since K has to be a positive quantity, that means that this quantity here approaches 0. So that means that applying the limit, this is going to be equal to. Hi. Multiplied by the quantity of 0, plus 1 divided by the quantity of 2K in quantity, since each of the zero is just 1. And simplifying this expression, we see that this is going to be equal to pi divided by the quantity of 2k in quantity. So next we need to find our volume B2. And V2 is going to be the volume of the solid that's formed by revolving our region R about the Y axis. So here we're actually going to use the shell method. And so we call for the shell method that V2 is going to be equal to. 2 i multiplied by the integral from 0 to infinity. Of the radius of our show, which is just going to be X. Multiplied by the height of our shell, which is just gonna be a value for Y, which is gonna be E, raised to the quantity of minus K multiplied by X and quantity DX. So, again, here we have an improper interval. So that means that we're gonna have to rewrite this as being equal to the limit, as R approaches infinity of 2 pi, multiplied by the integral from 0 to R of X, multiplied by E raises to the quantity of minus KX and quantity DX. So in order to perform this integration, we'll have to integrate by parts. So, to integrate by parts, we're going to set U equal to X, and that means that DV is going to be equal to E rates to the quantity of minus KX and quantity DX. So that means that DU is going to be equal to DX and that means that V is going to be equal to -1 divided by K multiplied by E raises the quantity of minus KX in quantity. So, performing the integration, this is going to be equal to the limit, as R approaches infinity. Of 2 multiplied by the quantity of, and here we call for integration parts, we'll have U multiplied by V, but we'll have minus X divided by K multiplied by E raised to the quantity of minus KX in quantity, and that needs to be evaluated from 0 to R. And then we'll have minus. The integral from 0 to R. Of VDU. So that's going to be the quantity of -1 divided by K, multiplied by E raised to the quantity of minus KX in quantity multiplied by DX in quantity. So, performing the second integration, we see that this is going to be equal to the limit, as R approaches infinity of 2 pi. Multiplied by the quantity, and here we'll substitute in our limits of integration for our boundary term. So we'll have minus R divided by K, multiplied by E raises to the quantity of minus KR in quantity. Plus 0 divided by K. Multiplied by E, raise the quantity of minus K multiplied by 0 in quantity. Then for the second integral, we'll factor out the constant of minus 1 divided by K, so we'll have plus 1 divided by K. Multiplied by the quantity, and here we integrate E rates to the quantity of minus KX in quantity, which will give us -1 divided by K multiplied by E. Rates of the quantity of minus KX in quantity, and this needs to be evaluated from 0 to R in quantity. So, simplifying this expression, this is going to be equal to the limit, as R approaches infinity of 2 pi. Multiplied by the quantity of minus R divided by K, multiplied by E raises the quantity of minus KR in quantity. Minus 1 divided by K2, multiplied by the quantity, and here we'll substitute in our limits of integration into E rates to the quantity of minus KX in quantity, and we'll get. E raised to the quantity of minus KR in quantity, minus E raises to the quantity of minus K multiplied by 0 in quantity. So now we need to take the limit as our approaches infinity. And if we look at our first term there, as our approaches infinity, R is going to approach infinity, but that gets multiplied by E goes to the minus KR which approaches 0 as R approaches infinity. And so what we need to do is apply Lopital's rule in order to figure out what value that limit actually approaches. So, we're gonna look at the limit. As R approaches infinity of are divided by. E raises the quantity of KR in quantity. And if we look at this case, then that means that both our numerator and denominator approach infinity as our approaches infinity. So we can apply Lopital's rule, and so we can take a derivative with respect to R both our numerator and denominator. So that means that this is going to be equal to the limit, as R approaches infinity of 1 divided by the quantity of 1 divided by K multiplied by E raises to the quantity of K R in quantity. And as R approaches infinity, that denominator is going to approach infinity. And so once you get 1 divided by infinity, which approaches 0. So that means that this limit is going to be equal to 0. That means the first term is going to be equal to 0, so that means that this is going to be equal to 2 pi, multiplied by the quantity of 0. -1 divided by K2d. Multiplied by the quantity, and the first term in this factor, which is E raises the quantity of minus KR is going to approach 0 as R approaches infinity, so we'll have 0, then we'll have minus 1. In quantity, and simplifying this expression, we see that this is just going to be equal to. To pie Divided by K squared. Now we need to see if these two volumes can be equal to each other for some value of K. So setting V1 equal to V2, we will have that. Pi divided by the quantity of UK. In the quantity, going to be equal to. To pie Divided by k squared. So we need to solve this for K. So dividing both sides by pi. And multiplying both sides by 2. And as well as multiplying both sides by K squared. We will get that K is equal to 4, and that is the answer that we're looking for for this problem. And if we compare that to our answer choices, we see that the correct answer choice corresponds to answer choice D. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Volumes of Solids of Revolution

Improper Integrals and Convergence

Function Behavior and Parameter Dependence

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