1. Let R be the region bounded by the x-axis, the graph of y=sqr(x) , and the line x=4 .

a. Find the area of the region 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 around the x – axis, they generate solids with equal volumes. Find the value of k.

a)

Area of region from x=0 to x=4
= ∫x^(1/2) dx from x=0 to 4
= [ (2/3)x^(3/2) ] from x = 0 to 4
= (2/3)(4^(3/2)
= (2/3)(8) = 16/3

b)
so we want
∫x^(1/2) dx from x = 0 to h to have a value of 8/3

(2/3)h^(3/2) = 8/3
h^(3/2) = (8/3)(3/2) = 4

h = 4^(2/3) = 16^(1/3) or cuberoot(16) or appr 2.52

c) this next part can be done in the same way
first let's find the volume from x-0 to 4

V = π∫ x dx from 0 to 4
= [(1/2)πx^2 ] from 0 to 4
= (1/2)π(16) = 8π

d) set the integral equal to 4π and solve for k
I am sure you can handle the rest

thanks

a. Well, to find the area of region R, we can integrate the function y=sqr(x) from x=0 to x=4. But wait, why did the mathematician never trust calculus teachers? Because they were always "integrating"! Anyway, let's calculate that area without any funny business.

The area of region R = ∫[0 to 4] sqr(x) dx

b. Now, 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 find the x-coordinate where the cumulative area of both regions is equal. It's like finding the balance point, but not for your bank account!

We need to solve the equation: ∫[0 to h] sqr(x) dx = ∫[h to 4] sqr(x) dx

c. To find the volume when region R is revolved about the x-axis, we need to use the disk method (no, not the kind you eat off of!). We can consider each infinitesimally small vertical strip of width dx as a disk and integrate their volumes. Just remember to be gentle, these disks have a lot of sensitive data!

The volume of the solid = ∫[0 to 4] π * sqr(x)^2 dx

d. Finally, to find the value of k such that the regions divided by the line x=k generate solids with equal volumes when revolved around the x-axis, we need to solve the equation:

∫[0 to k] π * sqr(x)^2 dx = ∫[k to 4] π * sqr(x)^2 dx

So, there you have it! Don't worry if it seems a bit confusing, math can be as tricky as an acrobatic clown. Just take your time and give it your best shot!

To solve this problem step-by-step:

a. To find the area of the region R, we need to integrate the function f(x) = sqrt(x) from x = 0 to x = 4. The area can be calculated using the definite integral:

Area = ∫[0, 4] sqrt(x) dx

Using the power rule of integration, we have:

Area = [(2/3) * x^(3/2)] [from 0 to 4]
= (2/3) * 4^(3/2) - (2/3) * 0^(3/2)
= (2/3) * 8 - (2/3) * 0
= 16/3

Therefore, the area of region R is 16/3 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 need to set up the following equation:

∫[0, h] sqrt(x) dx = ∫[h, 4] sqrt(x) dx

Using the same integration formula, we can rewrite the equation as:

(2/3) * h^(3/2) = (2/3) * 4^(3/2) - (2/3) * h^(3/2)

Simplifying the equation, we get:

h^(3/2) = 4^(3/2) - h^(3/2)

Taking the cube of both sides, we have:

h^3 = 64 - h^3

2h^3 = 64
h^3 = 32
h = ∛32
h ≈ 3.15

Therefore, the value of h such that the vertical line x = h divides the region R into two equal area regions is approximately 3.15.

c. To find the volume of the solid generated when region R is revolved about the x-axis, we will use the disk method. The volume can be calculated using the definite integral:

Volume = ∫[0, 4] π * (f(x))^2 dx

Using the function f(x) = sqrt(x), we have:

Volume = ∫[0, 4] π * (sqrt(x))^2 dx
= ∫[0, 4] π * x dx

Integrating, we get:

Volume = (π/2) * x^2 [from 0 to 4]
= (π/2) * 4^2 - (π/2) * 0^2
= (π/2) * 16 - (π/2) * 0
= 8π

Therefore, the volume of the solid generated when region R is revolved about the x-axis is 8π cubic units.

d. To find the value of k such that the vertical line x = k divides the region R into two regions such that when these two regions are revolved around the x-axis, they generate solids with equal volumes, we can set up the following equation:

∫[0, k] π * (sqrt(x))^2 dx = ∫[k, 4] π * (sqrt(x))^2 dx

Simplifying the equation, we get:

∫[0, k] π * x dx = ∫[k, 4] π * x dx

Using the same method as in part c, integrating both sides, we have:

(π/2) * k^2 = (π/2) * 4^2 - (π/2) * k^2

Simplifying the equation:

k^2 = 8 - k^2
2k^2 = 8
k^2 = 4
k = 2

Therefore, the value of k such that the vertical line x = k divides the region R into two regions that generate solids with equal volumes is 2.

a. To find the area of the region R, you can use the definite integral. The region R is bounded by the x-axis, the graph of y = sqrt(x), and the line x = 4. We can find the area by integrating the function sqrt(x) from x = 0 to x = 4.

∫[0,4] sqrt(x) dx

To integrate this, you can use the power rule for integration:

∫[0,4] x^(1/2) dx = (2/3)x^(3/2) evaluated from 0 to 4

[(2/3)(4)^(3/2)] - [(2/3)(0)^(3/2)]

=(2/3)(8) - 0

=16/3

So, the area of the region R is 16/3 square units.

b. To find the value of h such that the vertical line x = h divides the region R into two equal areas, we can set up an equation and solve for h. Since we want to divide the area in half, the areas on both sides of the line x = h should be equal.

We need to find the equation for the area of the region to the left of the line x = h. This can be found by integrating the function sqrt(x) from x = 0 to x = h:

∫[0,h] sqrt(x) dx

Now, we set up the equation for the area:

∫[0,h] sqrt(x) dx = ∫[h,4] sqrt(x) dx

We can solve this equation to find the value of h.

c. To find the volume of the solid generated when region R is revolved about the x-axis, we can use the volume of revolution formula. The volume can be found by integrating the area of the cross-sections perpendicular to the x-axis as we rotate the region.

V = ∫[0,4] π(sqrt(x))^2 dx

V = ∫[0,4] πx dx

V = π∫[0,4] x dx

Using the power rule for integration:

V = π(x^2/2) evaluated from 0 to 4

V = π(16/2) - π(0)

V = 8π

So, the volume of the solid generated is 8π cubic units.

d. To find the value of k such that the vertical line x = k divides the region R into two regions of equal volume, we can set up an equation and solve for k. The volumes on both sides of the line x = k should be equal.

We need to find the equation for the volume of the region to the left of the line x = k. This can be found by integrating the function π(sqrt(x))^2 from x = 0 to x = k:

∫[0,k] π(sqrt(x))^2 dx

Now, we set up the equation for the volume:

∫[0,k] π(sqrt(x))^2 dx = ∫[k,4] π(sqrt(x))^2 dx

We can solve this equation to find the value of k.