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: ∫[4,1] sqrt(1+(dy/dx)^2)
dy/dx = 1/2sqrt(x)
(dy/dx)^2 = 1/4x
1+ (dy/dx)^2 = (4x+1)/(4x)
∫[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
∫[16,1] --> is this simplified? Or can I say ∫[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))
∫[4,1] 2pi(sqrt(x+(1/4)))dx
B) Y-AXIS
R = y^2
ds = 1+4y^2 dy
∫[16,1] 2pi(sqrt(y^2+4y^2))dy --> Is this equal to ∫[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?
I would really appreciate if you could please help me out! Thank You!
NOTE!!
I just realized all of my integral symbols turned into a^<< please note, they are meant to be the integral symbol!
I know this is completely off topic...but do you watch Marvel's Agents of SHIELD??
#1 correct
∫[1,4] √(1 + 1/4x) dx
∫[1,2] √(1+4y^2) dy
where did you come up with [16,1]?
The curve goes from (1,1) to (4,2)
Note that √(1+4y^2) is NOT (1 + 2y)
√(a^2+b^2) is not a+b !!
#2. looks ok.
For C, the radius is just x+2
For D, the radius is just 3-y
Oh sorry about the that, I was starting to do u substitution and was changing my limits to [16,1] but then remembered I didn't have to evaluate.
And gotcha! so then for C would the set up be
∫[-2,4] 2pi(sqrt(x) - 2) √(1 + (dy/dx)^2) dx
oh wait... wouldn't x = -2 give you a vertical line meaning that it would be a y-integration? Or does that not matter for this question?
Also for part D, how did you get 3-y? why wouldn't it be 3-y^2? Because I thought we would have to have "y" when solving (similar to question 2 part B?
To find the length of the portion of the curve y = sqrt(x) between the lines x = 1 and x = 4, you need to set up two integrals, one with respect to x and one with respect to y.
1) Length of A with respect to x:
The formula for arc length is given by L = ∫[a,b] sqrt(1 + (dy/dx)^2) dx.
In this case, dy/dx = 1/(2*sqrt(x)). Now, let's substitute this into the formula:
L = ∫[4,1] sqrt(1 + (1/(2*sqrt(x)))^2) dx
Simplifying further:
L = ∫[4,1] sqrt(1 + 1/(4x)) dx
2) Length of A with respect to y:
The formula for arc length is given by L = ∫[c,d] sqrt(1 + (dx/dy)^2) dy.
In this case, x = y^2. Differentiating both sides with respect to y, we get dx/dy = 2y.
Now, let's substitute this into the formula:
L = ∫[c,d] sqrt(1 + (2y)^2) dy
Simplifying further:
L = ∫[16,1] sqrt(1 + 4y^2) dy
For the area of the surfaces generated by revolving the curve C about different axes, we can use similar approaches:
A) Revolving about the X-axis:
The formula for surface area is given by A = ∫[a,b] 2πy sqrt(1 + (dy/dx)^2) dx.
In this case, y = sqrt(x) and dy/dx = 1/(2*sqrt(x)). Now, let's substitute this into the formula:
A = ∫[4,1] 2π(sqrt(x)) sqrt(1 + (1/(2*sqrt(x)))^2) dx
Simplifying further:
A = ∫[4,1] 2π(sqrt(x)) sqrt(1 + 1/(4x)) dx
B) Revolving about the Y-axis:
The formula for surface area is given by A = ∫[c,d] 2πx sqrt(1 + (dx/dy)^2) dy.
In this case, x = y^2 and dx/dy = 2y. Now, let's substitute this into the formula:
A = ∫[d,c] 2π(y^2) sqrt(1 + (2y)^2) dy
Simplifying further:
A = ∫[16,1] 2π(y^2) sqrt(1 + 4y^2) dy
For parts C and D, you would indeed need to use the washer or shell method, depending on how the curve is being revolved. I would need more information to help you with parts C and D.
I hope this helps! Let me know if you have any further questions.