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 = π/4 and x = 5π/4. The cross-sections between these planes are circular disks whose diameters run from the curve y = 2 cos x to the curve y = 2 sin 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. Find the volume of the solid bounded by planes perpendicular to the x axis at x is equal to pi divided by 4 and x is equal to 5 pi divided by 4. Whose cross sections perpendicular to the x axis are circular discs with diameters joining the curves, y is equal to 4 multiplied by cosine of x and y is equal to 4 multiplied by sin of x. OK, so it appears for this particular prong, we're asked to take all the information that is provided to us by the prom itself, and we're asked to ultimately determine what the volume of this solid is. So now that we know that we're ultimately solving for what the volume of this solid is, we must first recall, based on our prior knowledge, what the equation is for the volume of a solid, specifically the volume of a solid of an integrable, meaning we can integrate cross-sectional area a of x on the closed interval square brackets of A, B. And this formula for the volume is written as V the volume is equal to the integral evaluated from A to B of AXDX. Fantastic. And we need to recall a note where the area a of x is equal to pi multiplied by parentheses of r x to the 2. Is going to be our expression for the area for the circular discs. So the area of the circular discs is equal to A of X. So now, at this stage, we need to determine the diameter and radius of each disk, and we need to note that the prom itself tells us that 4 multiplied by sin of x will be greater than or equal to 4 multiplied by cosine of x on the closed interval pi divided by 4. 5 pi divided by 4. Fantastic. So, this is based on the information that is given to us by the prom itself, and based on this information, we could say that 4 multiplied by sin of x is greater than or equal to 4 multiplied by cosine of x on this specific closed interval of 4 divided by pi, 5 pi divided by 4. Thus, with this in mind, the diameter, which we'll write as d of x, so the function d of x, the diameter expression, is given by, so d x is equal to 4 multiplied by sin of x minus 4 multiplied by cosine of x. That will mean then that r x, the radius function, will be equal to 1/2 multiplied by parentheses of. 4 multiplied by sin of x minus 4 multiplied by cosine of x, which will be equal to 2 multiplied by sin of x minus 2 multiplied by cosine of x. Fantastic. So. With that said, we can now solve for the area of the disk. So A of x is equal to, once again, pi multiplied by parentheses of R x2. So that means that it will be equal to pi multiplied by parentheses of 2 multiplied by sin of x minus 2 multiplied by cosine. Of x to the 2, and this will be equal to pi multiplied by parentheses of 4 multiplied by sin square of x minus 8 multiplied by sin of x multiplied by cosine of x plus 4 multiplied by cosine squared of x. Fantastic, and this will be equal to 4 pi multiplied by parentheses of sin squared of x. -2 multiplied by sin of x multiplied by cosine of x plus cosine square of x. And this will be equal to 4 pi multiplied by parentheses of 1 minus sin to the power of 2. Or rather it's gonna be 1 minus sin of 2 X. Fantastic. So, let's make a note. So we need to note that sin to the power of 2 x plus cosine to the power of 2 X is equal to 1. And we need to note that 2 multiplied by sin of x multiplied by cosine of x will be equal to sin of 2x. Fantastic. So now we could start to solve for our volume of the solid. So we could write that V, the volume is equal to the integral evaluated from pi divided by 4 to i pi divided by 4. Of 4 pi multiplied by parentheses of 1 minus sin of 2 x dx will be equal to. 4 pi multiplied by square brackets of the integral evaluated from pi divided by 4 to pi pi divided by 4 of d x minus the integral evaluated from pi divided by 4 to pi pi divided by 4 of sin of 2 x d x in square brackets. Will be equal to 4 pi multiplied by square brackets of the integral evaluated from 4 pi to 5 pi divided by 4 of DX. Minus 1/2 multiplied by the integral evaluated from pi divided by 4 to 5 pi divided by 4 of sin of 2 x multiplied by d of 2x. Will be equal to 4 pi multiplied by square brackets of x plus 1/2 multiplied by cosine of 2 x evaluated from pi divided by 4 to 5 pi divided by 4. OK. So at this stage, let us note that. At, so let's make a note that at X is equal to 5 pi divided by 4, that will mean that if we plug in a value of 5 pi divided by 4 in for x, we will get 5 pi divided by 4. Plus. 1 divided by 2 multiplied by cosine of 2 multiplied by 5 pi divided by 4, and when we simplify this expression, you will find that it's simply equal to 5 pi divided by 4. Fantastic. And at, so let us note that at x is equal to. Pi divided by 4, that will mean when we plug in a value of pi divided by 4 in for x, we will find that we can go ahead and write. That pi divided by 4 plus 1 divided by 2 multiplied by cosine of. 2 multiplied by pi divided by 4 will be equal to when we plug our expression into our calculator and we write it in its most simplified form, we will get pi divided by 4. Fantastic. So, therefore, our volume V will be equal to 4 pi multiplied by square brackets of, so square brackets of. 5 pi divided by 4 minus pi divided by 4, which will be simply equal to. Or pi to the power of 2. So 4 pi to the power 2 and our units will be cubic units, and this is our final answer. Hooray, we did it. So going back to the top to recap what we've learned, we can safely say without a doubt that our final answer for the volume of this solid will be once again 4 multiplied by pi to the power of 2 cubic units, and that's it. 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 = π/4 and x = 5π/4. The cross-sections between these planes are circular disks whose diameters run from the curve y = 2 cos x to the curve y = 2 sin 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 = π/4 and x = 5π/4. The cross-sections between these planes are circular disks whose diameters run from the curve y = 2 cos x to the curve y = 2 sin x.