Let A denote the portion of the curve y = sqrt(x) that is between the lines x = 1 and x = 4.

1) Set up, don't evaluate, 2 integrals, one in the variable x and one in the variable y, for the length of A.

My Work:
for x: integral[4,1] sqrt(1+(dy/dx)^2)
dy/dx = 1/2sqrt(x)
(dy/dx)^2 = 1/4x
1+ (dy/dx)^2 = (4x+1)/(4x)
integral[4,1] sqrt((1+1/4x)dx --> would this count for setting up? Or do I need to get rid of the sqrt? If I have to get rid of the sqrt, I am confused on how I would go about doing that.

for y: y^2 = x
dx/dy = 2y
(dx/dy)^2 = 4y^2
1 + (dx/dy^2) = 1 + 4y^2
sqrt(1 + (dx/dy^2)) = 1+4y^2
integral[16,1] --> is this simplified? Or can I say integral[16,1] 1+2y dy?

2) Set up, don't evaluate,
Set up, but do not evaluate, two integrals, one in the variable x and one in the variable y, for the area of the surfaces generated by revolving C about:

A) X-AXIS
R = sqrt(x)
ds = sqrt(1+(1/4x))
integral[4,1] 2pi(sqrt(x+(1/4)))dx

B) Y-AXIS
R = y^2
ds = 1+4y^2 dy
integral[16,1] 2pi(sqrt(y^2+4y^2))dy --> Is this equal to integral[16,1] 2pi(y+2y)dy

C) LINE X = -2
D) LINE Y = 3
I am confused on how I would do parts C and D. Am I supposed to incorporate the washer/shell method?

To set up the integrals for the length of curve A, you have correctly determined the values of dy/dx and (dy/dx)^2. However, in both cases, you need to simplify the expression inside the square root before setting up the integral.

For the integral in terms of x:
The expression inside the square root is (4x+1)/(4x). To simplify it, you can multiply the numerator and denominator by x to get (4x^2+x)/(4x). Simplifying further, you have (4x^2 + x)/(4x) = (4x^2/4x) + (x/4x) = x + 1/4.

So the integral in terms of x becomes:
∫[4,1] √(1 + 1/4) dx
This simplifies to:
∫[4,1] √(5/4) dx

For the integral in terms of y:
You correctly determined that y^2 = x, so x = y^2. However, when you differentiate x = y^2 with respect to y to find dx/dy, you are missing the chain rule.

Differentiating x = y^2 with respect to y:
dx/dy = 2y * dy/dy
Simplifying, we have dx/dy = 2y.

Now, for the integral in terms of y:
The expression inside the square root is (1 + (dx/dy)^2) = 1 + (2y)^2 = 1 + 4y^2.

So the integral in terms of y becomes:
∫[16,1] √(1 + 4y^2) dy

Now let's move on to setting up the integrals for the area of the surfaces generated by revolving curve C.

A) Revolving C about the X-axis:
The expression for the surface area element ds is correct: √(1 + (1/4x)).

To set up the integral in terms of x:
The integral represents the sum of the surface area elements as x varies from 1 to 4. So the integral becomes:
∫[4,1] 2π(sqrt(x + 1/4)) dx

B) Revolving C about the Y-axis:
The expression for the surface area element ds is incorrect. It should be √(1 + (dx/dy)^2), which simplifies to √(1 + 4y^2).

To set up the integral in terms of y:
The integral represents the sum of the surface area elements as y varies from 1 to 16. So the integral becomes:
∫[16,1] 2π(sqrt(1 + 4y^2)) dy

C) Revolving C about the line X = -2:
To find the integral for this case, you can use the washer method. The integral represents the sum of the areas of the washer-shaped slices as x varies from 1 to 4.

You'll need to calculate the outer radius and the inner radius for each slice. The outer radius is the distance from the curve to the line X = -2, so it is x - (-2) = x + 2. The inner radius is the distance from the curve to the line X = 1, so it is x - 1.

The differential area is given by dA = π(R_outer^2 - R_inner^2) dx.

Thus, the integral becomes:
∫[4,1] π((x + 2)^2 - (x - 1)^2) dx

D) Revolving C about the line Y = 3:
To find the integral for this case, you can again use the washer method. The integral represents the sum of the areas of the washer-shaped slices as y varies from 1 to 16.

You'll need to calculate the outer radius and the inner radius for each slice. The outer radius is the distance from the curve to the line Y = 3, so it is y - 3. The inner radius is the distance from the curve to the line Y = sqrt(1), which is just y.

The differential area is given by dA = π(R_outer^2 - R_inner^2) dy.

Thus, the integral becomes:
∫[16,1] π((y - 3)^2 - y^2) dy

Remember that these integrals only represent the setup, and you still need to evaluate them to find the actual length and area.