The region between the curve y=2xy=\frac{2}{\sqrt{x}} and the xx-axis from x=1x=1 to x=3x=3 is revolved about the xx-axis to form a solid. Find the volume of this solid.

4 π ln ⁡ ( 3 ) ≈ 13.81 4\pi\ln\left(3\right)\thickapprox13.81

The region between the curve y = 2 x y=\frac{2}{\sqrt{x}} and the x x -axis from x = 1 x=1 to x = 3 x=3 is revolved about the x x -axis to form a solid. Find the volume of this solid.

this problem, we're told that the region between the curve y equals two over root x and the x axis from x equals one to x equals three is revolved about the x axis to form a solid. And we're asked here to find the volume of this solid. So we're gonna be working with a solid of revolution here. Whenever we work with these solids, we wanna go ahead and sketch them just so we have a good idea of what we're dealing with and how we should ultimately set up our volume integral. So looking at this function two over root x, I really just need a rough sketch of what the shape of this looks like. And two over root x looks something like this. Now if I take this function from x equals one to x equals three, just marking those points on my axis here, and I'm gonna go ahead and draw some dotted lines just to show what area we're working with. We're gonna take this area. So the region between our curve and the x axis from x equals one to x equals three, that's what our problem tells us. We're taking that area. We're revolving it around that x axis. So I'm gonna draw the other side of my solid here as we revolve that around. We're gonna kind of mirror the other side of that solid. And now we can kind of see what solid of revolution we are working with. Now based on this solid, I also wanna go ahead and sketch its cross section. Here, we can see it's going to have a circular cross section as I've taken that area and I have revolved it in a circular motion around that x axis. So looking at a circular cross section here, we wanna focus on what that radius is. Now looking at my graph here, I can see that my radius goes from my curve to my axis. Axis. So that radius, r of x, is just going to be equal to my function minus my axis. So that's two over root x minus zero, which is just my function itself because that minus zero doesn't really do anything there. So now that I know what the radius of a cross section out of the solid is, I can go ahead and set up my volume integral. Remember that when using the disk method working with a solid of revolution, our volume is going to be equal to the integral from a to b of pi r squared dx, where r is a function of x since here we were revolving around that x axis. So setting up my volume integral here, this is the integral from one to three of pi times my radius, which is that two over root x squared, dx. Now I'm going to go ahead and square that two over root x to give me pi times four over x dx, and that is again the definite integral from one to three. Now I'm gonna go ahead and pull both my constants out, that four and that pi. So that's this is just four pi times the integral from one to three of one over x dx. Now the integral of one over x dx is just the natural log of the absolute value of x. So that's what this gives me here, four pi times the natural log of the absolute value of x. And then we're bounded here from one to three. So now let's go ahead and plug in our bounds. So this is four pi. I'm gonna go ahead and keep my constants out front there. Natural log of the absolute value of three, which is just natural log of three minus natural log of one. Those are both already positive values, so I don't have to worry about that absolute value. Now the natural log of one is just zero, so I'm really left here with four pi times the natural log of three. Now this is an appropriate answer, but if you are using a calculator and you want this value as a decimal, you can just type this right into your calculator and you will ultimately get here 13.81 having rounded that to two decimal places. So remember, this isn't just some random value. It represents the volume of our solid of revolution here. Feel Feel free to let us know if you have any questions in the comments, and I'll see you in the next one.

The region between the curve y = 2 x y=\frac{2}{\sqrt{x}} and the x x -axis from x = 1 x=1 to x = 3 x=3 is revolved about the x x -axis to form a solid. Find the volume of this solid.

4 π ln ⁡ ( 3 ) ≈ 13.81 4\pi\ln\left(3\right)\thickapprox13.81

8 π ln ⁡ ( 3 ) ≈ 27.61 8\pi\ln\left(3\right)\thickapprox27.61

4 π ≈ 12.56 4\pi\thickapprox12.56

2 π ln ⁡ ( 3 ) ≈ 6.90 2\pi\ln\left(3\right)\thickapprox6.90

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Logarithmic Functions

Calculus

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Logarithmic Functions

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Multiple Choice

The region between the curve $y = \frac{2}{\sqrt{x}}$ and the $x$ -axis from $x equals 1$ to $x equals 3$ is revolved about the $x$ -axis to form a solid. Find the volume of this solid.

$8 pi l n of 3 almost equals 27.61$

$4 pi almost equals 12.56$

$2 pi l n of 3 almost equals 6.90$

$4 pi l n of 3 almost equals 13.81$

Verified step by step guidance

Step 1: Recognize that the problem involves finding the volume of a solid of revolution. The formula for the volume of a solid obtained by revolving a region around the x-axis is given by: V = ∫[a, b] π * (f(x))^2 dx, where f(x) is the function describing the curve, and [a, b] is the interval of integration.

Step 2: Identify the function and the interval. Here, the curve is y = 2/sqrt(x), and the interval is from x = 1 to x = 3. Substitute f(x) = 2/sqrt(x) into the volume formula.

Step 3: Set up the integral for the volume: V = ∫[1, 3] π * (2/sqrt(x))^2 dx. Simplify the integrand: (2/sqrt(x))^2 = 4/x. The integral becomes V = ∫[1, 3] π * (4/x) dx.

Step 4: Factor out constants from the integral: V = 4π ∫[1, 3] (1/x) dx. The integral of 1/x is ln|x|, so the integral becomes V = 4π [ln(x)] evaluated from x = 1 to x = 3.

Step 5: Apply the limits of integration: V = 4π [ln(3) - ln(1)]. Since ln(1) = 0, the result simplifies to V = 4π ln(3). This is the volume of the solid.

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