Volumes
Find the volumes of the solids in Exercises 1–18...

Question

Volumes

Find the volumes of the solids in Exercises 1–18.

The solid lies between planes perpendicular to the x-axis at x = 0 and x = 1. The cross-sections perpendicular to the x-axis between these planes are circular disks whose diameters run from the parabola y = x² to the parabola y = √x.

Accepted Answer

Hello there, today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A solid lies between planes perpendicular to the x axis at x equals 0 and x equals 2. The cross sections perpendicular to the x axis are circular disks whose diameters run from the curve y equals x to the power of 2 to the line y equals 2 x. Find the volume of the solid. OK, so it appears for this particular problem, based on all the information that is provided to us, we're asked to ultimately find what the volume of the solid is. So now that we know that we're ultimately solving for the volume of this solid. Our first step that we need to take is we need to recall and use the equation to help us solve for the volume of a solid, specifically the volume of a solid of an integrable, meaning it could be integrated, cross-sectional area A of X. On a closed interval, square brackets of A, B, which we can write this formula as V the volume is equal to the integral evaluated from A to B of AXDX. Where our area function A of X is equal to pi multiplied by parentheses of R X to the power of 2, and this is the area for the circular disks. Now we need to solve for the diameter and radius of each disk. We need to recall a note based on what the problem itself tells us that the diameter runs from y equal to x 2 to y equal to 2 x. Thus we could write that our diameter function d x is equal to 2 x minus x 2. Therefore, that would mean that R X, the radius function RX, will be equal to 2x minus x 2, all divided by 2. We can now solve for the area of the disk, so we can state that a of x is equal to pi multiplied by parentheses of r x to the power of 2, which will be equal to pi multiplied by parentheses of 2 x minus x to the 2 divided by 2 all to the power of 2. Which will be equal to pi divided by 4 multiplied by parentheses of 4 x 2 minus 4 x 3 plus x the power of 4. Fantastic So now we need to integrate from X is equal to 0 to X is equal to 2, given that the bounds and the curves meet at, so they're going to be meeting at X equals 0 and X equals 2. So considering this. We can go ahead and write that the volume b is equal to the integral evaluated from 0 to 2 of pi divided by 4 multiplied by parentheses of 4 multiplied by x to the 2 minus 4 multiplied by x 3 plus x 4. D x, which will be equal to pi divided by 4 multiplied by parentheses of 4 divided by 3 multiplied by x to the power of 3 minus 4 divided by 4 multiplied by x to the power of 4 plus 1 divided by 5 multiplied by x to the power of 5 evaluated from 0 to 2. And this will be equal to pi divided by 4 multiplied by parentheses of 4 divided by 3 multiplied by x to the power of 3 minus x 4 plus 1 divided by 5 multiplied by x to the power of 5 evaluated from 0 to 2. We also need to note that at, so let's make a note that at. X equals 2. So once again, if we're plugging in a value of 2 in for X, you'll be able to write that 4 divided by 3 multiplied by 0 to the power, or actually we're plugging in a value of 2, so it'd be 4 divided by 3. Multiplied by 23 minus 2 to 4. Plus 1 divided by 5 multiplied by 2 to the power of 5, which when we plug this expression into our calculator and we write our answer as a simplified fraction or simplified to its most simple form, we will find that it's equal to 16 divided by 15. And when At x is equal to 0, we could plug in a value of 0 in for x. We'll find that 4 divided by 3 multiplied by 0 to the power of 3 minus 0 to the power of 4 plus 1 divided by 5 multiplied by 0 to the power of 5 will be simply equal to 0. That will mean therefore that the volume v will be equal to pi divided by 4 multiplied by parentheses of 16 divided by 15 minus 0, which will be equal to 4 pi divided by 15. Cubic units when we write our final answer in its most simplified form and that's it that is our final answer hooray we did it so looking at the problem itself we could safely say that the volume of this solid once again is without a doubt. 4 pi divided by 15 cubic units, and that's it, that is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

VolumesFind the volumes of the solids in Exercises 1–18.The solid lies between planes perpendicular to the x-axis at x = 0 and x = 1. The cross-sections perpendicular to the x-axis between these planes are circular disks whose diameters run from the parabola y = x² to the parabola y = √x.

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Volumes
Find the volumes of the solids in Exercises 1–18.
The solid lies between planes perpendicular to the x-axis at x = 0 and x = 1. The cross-sections perpendicular to the x-axis between these planes are circular disks whose diameters run from the parabola y = x² to the parabola y = √x.