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) B. Find the value of h, such that the vertical line x = h divides the region R into two Regions of equal area.
A = ∫[1,3] 1/x^3 dx
= -1/2x^2 [1,3]
= 4/9
Now, we want to find c such that
∫[1,c] 1/x^3 dx = ∫[c,3] 1/x^3 dx
(-1/2c^2 - (-1/2)) = (-1/18 - (-1/2c^2))
(c^2-1)/2c^2 = (9-c^2)/18c^2
c = 3/√5
Thank you!!
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) To find the area of region R, we can set up an integral. Since region R is bounded by the x-axis and the curves x = 1, x = 3, and y = 1/x^3, we need to find the bounds of integration and the integrand.
1) Determine the bounds of integration:
The region R is bounded by the vertical lines x = 1 and x = 3. Therefore, the bounds of integration for x should be from 1 to 3.
2) Determine the integrand:
The region R is also bounded by the curve y = 1/x^3. This implies that the height of the region at any given x-value will be 1/x^3. The integrand will be the function that represents the height of the region at each point, which is 1/x^3.
3) Set up and evaluate the integral:
The area of region R can be calculated using the definite integral:
A = ∫[1 to 3] (1/x^3) dx
To evaluate this integral, we can use the power rule of integration. Recall that the integral of x^n dx is (1/(n+1)) * x^(n+1). Applying this rule, we get:
A = [-1/(3x^2)] evaluated from 1 to 3
Evaluating this expression gives:
A = [-1/(3(3^2))] - [-1/(3(1^2))]
A = -1/27 + 1/3
A = 8/27
Therefore, the area of region R is 8/27 square units.
B) To find the value of h such that the vertical line x = h divides region R into two regions of equal area, we need to set up and solve an equation representing the equation of area.
Let's denote the left region's area as A1 and the right region's area as A2. Since we want to divide the region into two equal parts, A1 = A2.
1) Determine the bounds of integration:
We need to find the bounds of integration for the left and right regions.
For the left region, since it is bounded by the x-axis, the curve y = 1/x^3, and the vertical line x = h, the bounds of integration for x should be from 1 to h. Denote this integration as I1.
For the right region, since it is bounded by the x-axis, the curve y = 1/x^3, and the vertical line x = h, the bounds of integration for x should be from h to 3. Denote this integration as I2.
2) Set up the equation of area:
Since A1 = A2 and each area is represented by an integral, we need to set up the equation:
I1 = I2
∫[1 to h] (1/x^3) dx = ∫[h to 3] (1/x^3) dx
3) Solve for h:
To solve this equation, we can evaluate the integrals and equate them.
∫[1 to h] (1/x^3) dx = ∫[h to 3] (1/x^3) dx
[-1/(3x^2)] evaluated from 1 to h = [-1/(3x^2)] evaluated from h to 3
[-1/(3h^2)] - [-1/(3(1^2))] = [-1/(3(3^2))] - [-1/(3h^2)]
Simplifying the equation gives:
-1/(3h^2) + 1/3 = -1/27 + 1/(9h^2)
To solve for h, we can clear the fractions and rearrange the terms:
9h^2 - 3 = h^2 - 1
8h^2 = 2
h^2 = 2/8
h^2 = 1/4
Taking the square root of both sides gives:
h = 1/2
So, the vertical line x = 1/2 divides region R into two regions of equal area.