42-47. Volumes of Solids Find the volume of the solid generate...

Question

42-47. Volumes of Solids Find the volume of the solid generated when the given region is revolved as described.

42. The region bounded by f(x) = ln(x), y = 1, and the coordinate axes is revolved about the x-axis.

Accepted Answer

Hello. In this video, we are going to be finding the volume of the solid generated when the region bounded by F of X equals to the natural logarithm of X and Y is equal to 2, and the coordinate axis are revolved about the y axis. Now, in order to approach this problem, the first thing we are going to do is we are going to draw the giving curves. Now, the first curve that is given to us is F of X is equal to the natural logarithm of X. Now this graph has an X intercept at the point 1.0. It infinitely approaches the y axis because it is not defined at zero and increases exponentially over time. So this is going to be the shape of the curve. Y is equal to LN of X. Now we are given a second curve, which is Y equal to 2. This is the equation of a horizontal line. Y equal to 2. And finally we are told that this region is bounded by the coordinate axises. Now, we need to go ahead and find the volume when the shape is revolved about the X axis. Now, because we are revolving about the X axis, we could integrate the shape with respect to X. Now, in this case here, because the natural logarithm of X is not defined as it approaches 0, we only want to integrate about defined values of X. So what that means is that we only want to integrate from 1 to 2, or from 1 to this point of intersection here. So the shape that we are going to be revolving is going to be this shaded region here. Now, in order to approach this problem, if we revolve about the X axis, we can go ahead and approach this by using the disk method. Now, the DC method is going to allow us to define the volume as pi multiplied by an integral from A to B, of a radius squared DX. Now, in this case here, we need to go ahead and find the bounds of integration. Now, because the natural logarithm function has a defined value of X at 1, the lowest boundary is going to be X is equal to 1. But the highest value of X is going to be defined at the point of intersection between L of X and Y is equal to 2. In order to find the X value of this region, we are going to set both of the equations equal to each other. That is going to give us the natural logarithm of X equal to 2. Now, we need to remove this X from the natural logarithm function. In order to do this, we can go ahead and exponentiate both sides of this equation, and that is going to allow us to simplify the equation to be X is equal to E to the power of 2. This is going to be the largest value of X at the point of intersection. So our bounds of integration are going to be from 1 to E squared. Next, what we need to do is we need to define our radius for the disk method. Now, the radius is going to be the distance from the X-axis to the uppermost curve of the shaded region. And again, this curve is defined by the natural logarithm of X, so our radius is going to be defined as the natural logarithm of X. So now that we have the pieces we need, we can go ahead and create an integral that will give us the volume for the shaded region. So the volume is going to be defined as pi multiplied by the integral from 1 to E squared of the natural logarithm squared of XDX. Now, in order to approach this kind of integral, we need to go ahead and use integration by parts. We are going to allow you to equal to the natural logarithm squared of X, and by taking the derivative of both sides of this equation, D U will equal to 2 multiplied by the natural logarithm of X divided by X. Next, we will go ahead and define our DV function to be DX and by integrating both sides of this equation, we will give V is equal to X. So now by putting together the integration by parts, we will be left with i. Multiplied by the quantity, V multiplied by X, which is going to be X. Multiplied by the natural logarithm squared of X, and this will be evaluated from 1 to E squared. Minus the integral from 1 to E squared, a V multiplied by DU, which is going to be 2 X multiplied by the natural logarithm of X, divided by XDX. Now there's a few things we need to do in order to simplify this equation. Or in order to simplify this antiderivative. The first thing we can go ahead and do is we can eliminate the X from both the numerator and denominator of the integral. And we can move this to to the front of the integral. So that is going to allow us to rewrite the antiderivative aspi multiplied by X L 2 X evaluated from 1 to E2 minus 2 multiplied by the integral from 1 to E2d of the natural logarithm of XDX. Now, we can simplify this integral by using an identity for the integral of Alan of X. The integral of Alan of X is defined as X multiplied by the natural logarithm of X minus X. And so what this means is that this part of the integral is going to simplify to be negative2 multiplied by the quantity X multiplied by the natural logarithm natural logarithm of X minus X. And by distributing this negative 2 into the quantity, we will have 2 XLan of X. Plus 2 X. So what this means is that this entire antiderivative is going to simplify to be pi multiplied by the quantity, XL square of X. -2 XLen of X. Plus 2 X and this entire thing will be evaluated from 1 to E squared. And so now what we are going to do is we are going to use the fundamental theorem of calculus part 2, and plug in the quantities. So by plugging in these quantities, we will get i multiplied. By the quantity, E2, multiplied by the natural logarithm squared of Esquared. -2 multiplied by E2, multiplied by the natural logarithm of E2. Plus 2 e squared. Then we will subtract the value of one plugged into the antiderivative, and that is going to be Ellen squared of 1. -2 Allan of 1. Plus 2 multiplied by 1 which is 2. Now there are a few things to simplify here. First, Allan of one will simplified to be zero. And we can rewrite Allen of E squared to be 2 multiplied by Ein of E. The same thing can be applied here, and we will have the quantity to. Allen of E quantity squared. And so what this is going to simplify to be is all of this will simplify to be pi. Multiplied by 2 E 2 minus 2, and by factoring out a 2, we will be left with 2 pi multiplied by the quantity E2 minus 1, which is going to be the volume of the region that is revolved about the X axis. And with that being said, the overall solution to this problem is going to be C. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

42-47. Volumes of Solids Find the volume of the solid generated when the given region is revolved as described.42. The region bounded by f(x) = ln(x), y = 1, and the coordinate axes is revolved about the x-axis.

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42-47. Volumes of Solids Find the volume of the solid generated when the given region is revolved as described.
42. The region bounded by f(x) = ln(x), y = 1, and the coordinate axes is revolved about the x-axis.