Question

There are four parts to this one question, and would really appreciate if you could show and explain how you get to the answer, because I tried looking up how to find the answer myself, but nothing made sense. Thank you!

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


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

int x^-3 dx = -.5 x^-2 + c
at x = 3 = -.5/9
at x = 1 = -.5
so A = .5 - .5/9 = .5(8/9) = 4/9

B at x = h int = -.5/h^2
right half = -.5/9 +.5/h^2
left half = -.5 h^2 +.5
so
-.5 h^2 + .5 = -5/9 +.5/h^2
1/h^2 = .5 + 5/9 = 4.5/9 + 5/9 = 9.5/9 = 19/18

int 1 to 3 of pi (x^-6)dx
= -pi/5 x^-5 + c
at 3 =
etc etc etc