Question

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

A) Find the area of R - I got 4/9

B) B. Find the value of h, such that the vertical line x = h divides the region R into two Regions of equal area. - I got 3/√5

C. Find the volume of the solid generated when R is revolved about the x-axis. - idk(maybe 242pi/1215)

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. - idk

Answer 1

a) Area = ∫ 1/x^3 dx from x = 1 to 3
= [ -1/(2x^2) ] from 1 to 3
= -1/18 - (-1/2)
= 1/2 - 1/18 = 4/9
you had that, good job

b) ∫ 1/x^3 dx from 1 to h = ∫ 1/x^3 dx from h to 3
[-1/2x^2] from 1 to h = [-1/2x^2] from h to 3
-1/2h^2 + 1/2 = -1/18 + 1/2h^2
times 18h^2
-9 + 9h^2 = -h^2 + 9
10h^2 = 18
h^2 = 9/5
h = 3/√5 , heh, you are doing great.

c) vol = π∫ y^2 dx
= π ∫ 1/x^6 dx from 1 to 3
= π [ -1/(5x^5) ] from 1 to 3
= π (-1/1215 - (-1/5) )
= π (242/1215)
you had that also, wow!

d) let's do it the same way we did b)
π∫ 1/x^6 dx from 1 to k = π∫ 1/x^6 dx from k to 3
divide out the π
[ -1/(5x^5] from 1 to k = [ -1/(5x^5} from k to 3
-1/(5k^5) + 1/5 = -1/1215 + 1/(5k^5)
2/5k^5 = 244/1215
1/5k^5 = 122/1215
1/k^5 = 122/243
122k^5 = 243
k^5 = 243/122
k = 3/122^(1/5) = appr 1.148

testing with Wolfram:
www.wolframalpha.com/input/?i=%E2%88%AB1%2Fx%5E6+dx+from+1+to+1.147755+

www.wolframalpha.com/input/?i=%E2%88%AB1%2Fx%5E6+dx+from+1.147755+to+3

same volume for both, my answer is correct