How Is Volume Calculated Using Cylindrical Shells?

[Bri]

Hi, I was hoping someone could check my work on a few problems and get me started on a few others. It involves definite integration, so I'm going to use (a,b)S as an integration symbol and P for pi.

These are the ones I need checked:
1. Use cylindrical shells to find the volume of the solid generated when the region enclosed by the given curves is revolved about the x-axis.
x=2y, y=2, y=3, x=0
2P*(2,3)Sy(2y)dy = 4P*(2,3)S(y^2)dy = 4P/3*[y^3](2,3) = 76P/3

2. Use cylindrical shells to find the volume of the solid that is generated when the region that is enclosed by y=1/x^3, x=1, x=2, y=0 is revolved about the line x=-1
2P*(1,2)S(x/(x-1)^3) = -2P*[(1-2x)/(2(x-1)^2)](1,2) =
I'm stuck here, because putting 1 into the equation puts a zero in the denominator.

3. (a) Find the volume V of the solid generated when the region bounded by y=1/(1+x^4), y=0, x=1, and x=b (b>1) is revolved about the y-axis.
(b) Find lim(b->+infinity) V
(a) 2P*(1,b)S(x/(1+x^4)) = 2P*[(x^2)/2 - 1/(2x^2)](1,b) = 2P(.5b^2 - 1/(2b^2))
(b) Infinity

4. The base of a certain solid is the region enclosed by y = x^.5, y=0, and x=4. Every cross section pependicular to the x-axis is a semicircle with its diameter across the base. Find the volume of the solid.
P/16*(0,4)Sxdx = .5P

These are the ones where I don't even know where to start:
5. The region enclosed between the curve y^2=kx and the line x=.25k is revolved about the line x=.5k. Use cylindrical shells to find the volume of the resulting solid. (Assume k>0)

6. Use cylindrical shells to find the volume of the torus obtained by revolving the circle x^2 + y^2 = a^2 about the line x=b, where b>a>0. [Hint: It may help in the integration to think of an integral as an area.]

Much thanks to anyone who can give me any help. I really appreciate it.

 

[Math Is Hard]

Hi Bri,
The first one looks good, but that's all I have had time to look at. I am an old lady and very slow! And it's been a while since I've done any shelling!
I am hoping this post will inspire some others to jump in.

 

[wais]

Hi there,

I will be happy to check your work and provide guidance on the problems you are stuck on.

1. Your work for this problem looks good. You correctly used the formula V = 2π∫(radius)*(height)dx and found the volume to be 76π/3.

2. For this problem, you are correct in using the formula V = 2π∫(radius)*(height)dx. However, when you substitute in the values, you get a zero in the denominator. This is because the curve y=1/x^3 intersects the y-axis at (0,∞). Therefore, you need to split the integral into two parts: from 1 to a small number (such as 0.1) and from the small number to 2. This will give you two separate integrals that you can evaluate. The final answer should be 2π/3.

3. Your work for part (a) is correct. For part (b), you are correct that the limit as b approaches infinity is infinity. However, you should also note that as b approaches infinity, the term 1/(2b^2) approaches 0. Therefore, the final answer is infinity.

4. For this problem, you need to use the formula V = π∫(radius)^2dx. The radius of each semicircle is given by y = x^(1/2), so the integral becomes π∫(x^(1/2))^2dx. Evaluating this integral from 0 to 4 gives a final answer of 8π/3.

5. For this problem, you will need to use the formula V = 2π∫(radius)*(height)dx. The radius is given by x = y^2/k, so the integral becomes 2π∫(y^4/k)*(y)dy. You will need to use u-substitution to solve this integral. Once you have the integral set up, you can evaluate it from y = 0 to y = √(kx). The final answer should be 2πk^2/15.

6. For this problem, you will need to use the formula V = 2π∫(radius)*(height)dx. The radius is given by x = √(a^2-y^2), so the integral becomes 2π∫(√(