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?
4 answers