Finding volume
Let R be the "triangular" region in the f...

Question

Finding volume

Let R be the "triangular" region in the first quadrant that is bounded above by the line y = 1, below by the curve y = ln x, and on the left by the line x = 1.

Find the volume of the solid generated by revolving R about

a. the x-axis.

Accepted Answer

Consider the region S located in the first quadrant, bounded by the curve Y equals natural log of X from below, the horizontal line Y equals 2 from above, and the vertical line X equals 1 on the left. Find the volume of the solid form in the regions is revolved around the x axis. We have 4 possible answers, being 2 pi, E2 + 12, E2 minus 1, pi E + 4, or pie minus 1, all in cubic units. Now we know the bottom battery is like X, and the top is Y equals 2. So, Let's find our endpoints. We have Y equals log X and Y equals 2. We'll set these equal to each other to find our bounce. Na log x equals 2. Gives us X equals e to the 2nd. So, our region is bounded by X equals 1 and X equals E to the 2nd on the horizontal direction. So, we will use the washer method based on what we know here. We have the integral. A a beef. Of our outer radius 2 minus our inner radius squad. And we will use DX for this. There's also a pi in the front of our integral. You have pi integral from 1 to E2. Of our outer Now, our outer will be 2, because it is the highest or largest value of our function. So we have 22 minus log X2 DX. So, we can simplify this to give us pi integral from 1 to E2. Of 4 minus natural log of x all squared d x. Now, let's go ahead and solve our integral. We can separate this into two integrals because we have a subtraction between our two functions. We have pi. Integral from 1 to E2 or D X. Minus the integral 1 to E2 of natural log X2DX. Now we know the first part of this is the integral of 4, which is just 4 X. So if pi multiplied by 4 X from 1 to E2. Now we have to solve our other integral. This one is integration by parts problem. So, we will note this down here, the integral from 1 to E2 of natural log X2. You will be natural log X. Square. DV will just be DX, that makes V equals to X. And then we have DU. Which is just 2 natural log X divided by XDX. So now we can simplify this integral. We have our integral. Which is UV. Or X natural log X2. Minus the integral of V, which is X, and then DU multiplied by 2 natural log X divided by XDX. We can simplify this even further to get x natural logx squared. Minus the integral to natural log XDX. Now, we do know what the integral of natural log of X is, if we use integration by parts. The integral of natural log X is equal to X log X minus X + C. So, we get X log X2 minus 2 multiplied by X log X minus X + C. So now we can combine our previous integral with. Our current integral We get pi multiplied. By 4 x Minus X natural lax. Squared -2 X natural log X minus X. All from 1 to E2. Now we can just plug in our two bounds. We get pi Multiplied by. 2 e 2. -2 when we plug in our bounce. This gives us Tupai Multiplied by e2 minus 1. This will be in cubic units. That means the answer to our problem, if we look at our possible answers. Is answer B. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

Finding volumeLet R be the "triangular" region in the first quadrant that is bounded above by the line y = 1, below by the curve y = ln x, and on the left by the line x = 1.Find the volume of the solid generated by revolving R abouta. the x-axis.

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Finding volume
Let R be the "triangular" region in the first quadrant that is bounded above by the line y = 1, below by the curve y = ln x, and on the left by the line x = 1.
Find the volume of the solid generated by revolving R about
a. the x-axis.