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.
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
whoops, I spun around y axis
same idea though
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.
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
Responsible-Use-of-University-Computing-and-Network-Resources-Policy.pdf
I spun around the wrong axis. That is why :)
a.) To find the area of region R, we need to integrate the function y = 1/x^3 with respect to x over the bounds x = 1 to x = 3.
The area A of region R can be calculated using the definite integral:
A = ∫[1 to 3] (1/x^3) dx
To solve this, we'll use the power rule of integration. The antiderivative of x^n is (1/(n+1)) * x^(n+1), where n is any real number except -1. Applying this rule to our integral, we get:
A = ∫[1 to 3] (1/x^3) dx = (1/(1+3)) * x^(-3+1) | [1 to 3]
= (1/4) * [3^(-2) - 1^(-2)]
Simplifying further,
A = (1/4) * [1/9 - 1]
= (1/4) * [-8/9]
= -2/9
However, since area cannot be negative, we take the absolute value:
A = | -2/9 |
A = 2/9
Therefore, the area of region R is 2/9 square units.
b.) To find the value of h such that the vertical line x = h divides the region R into two regions of equal area, we can set up an equation:
∫[1 to h] (1/x^3) dx = ∫[h to 3] (1/x^3) dx
Using the same formula mentioned in part (a), we can write:
(1/4) * [h^(-2) - 1^(-2)] = (1/4) * [3^(-2) - h^(-2)]
Simplifying further,
h^(-2) - 1 = 3^(-2) - h^(-2)
2h^(-2) = 8/9
h^(-2) = 4/9
(h^(-2))^(-1) = (4/9)^(-1)
h^2 = 9/4
h = √(9/4)
h = 3/2
Therefore, the value of h that divides region R into two equal areas is h = 3/2.
c.) To find the volume of the solid generated when region R is revolved about the x-axis, we can use the disk method.
Consider an infinitesimally thin disk of radius y and width dx. The volume of this disk is given by dV = πy^2 dx.
Integrating this volume over the bounds x = 1 to x = 3, we can calculate the total volume V:
V = ∫[1 to 3] π(1/x^3)^2 dx
Simplifying,
V = π ∫[1 to 3] (1/x^6) dx = π * [-1/(5x^5)] | [1 to 3]
= π * [(-1/(5(3^5))) - (-1/(5(1^5)))]
= π * [(-1/405) + (1/5)]
= π * [(1 - 81)/2025]
= π * (-80/2025)
= -80π/2025
Taking the absolute value, we get:
V = 80π/2025
Therefore, the volume of the solid generated when region R is revolved about the x-axis is 80π/2025 cubic units.
d.) To find the value of k such that the vertical line x = k divides the region R into two regions that generate solids with equal volumes when revolved about the x-axis, we can set up the following equation:
∫[1 to k] π(1/x^3)^2 dx = ∫[k to 3] π(1/x^3)^2 dx
Using the same formula mentioned in part (c), we have:
π ∫[1 to k] (1/x^6) dx = π ∫[k to 3] (1/x^6) dx
Simplifying further,
[(-1/(5k^5)) - (-1/405)] = [(-1/(5(3^5))) - (-1/(5k^5))]
Simplifying and rearranging the equation,
1/(5k^5) + 1/(5(3^5)) = 1/405
Let's multiply both sides by 5k^5 * 405 to eliminate the denominators:
405 + 405k^5 = 5k^5 * 405
Dividing both sides by 405 and simplifying,
1 + k^5 = 5k^5
4k^5 = 1
k^5 = 1/4
k = 1^(1/5)
k = 1/√(2)
Therefore, the value of k that divides region R into two regions that each generate solids with equal volumes when revolved about the x-axis is k = 1/√(2).