The region R is bounded by the x-axis, x = 1, x = 3, and y = 1/x^3.

Question

a.) Find the area of R.

b.) Find the value of h, such that the vertical line x = h divides the region R into two Regions of equal area.

c.) Find the volume of the solid generated when R is revolved about the x-axis.

d.) The vertical line x = k divides the region R into two regions such that when these two regions are revolved about the x-axis, they generate solids with equal volumes. Find the value of k.

Answer 1

a)
integral x^-3 dx = -(1/2)x^-2 + c
from 1 to 3
-(1/2)(1/9) + (1/2)(1/1)
= 1/2 - 1/18
= 8/18
= 4/9
B)

-(1/18)+(1/2)(1/h^2) = -(1/2)(1/h^2) + (1/2)(1/1)
1/h^2 = 1/2 + 1/18 = 10/18 = 5/9
h^2 = 9/5
h = 3/sqrt 5

C) integrate 2 pi x (1/x^3)dx from x = 1 to x = 3
2 pi [ -(1/2)1/x^2 ] + c
- pi /x^2 + c
then do the same as parts A and B

Answer 2

whoops, I spun around y axis
same idea though

Answer 3

These explanations make sense. I got 4/9 on my own and I was able to use that info to solve part B. but I'm still caught up on part D.

Answer 4

I told you what to do in
https://www.jiskha.com/display.cgi?id=1524618205

btw, in Damon's formula he forgot the square of the radius

Answer 5

Responsible-Use-of-University-Computing-and-Network-Resources-Policy.pdf

Answer 6

I spun around the wrong axis. That is why :)