HELP?! curve length and surface area!any idea?which is what

[amay9210]

Consider the curve given by x(t) = (1 + t^1/2)2 and y(t) = (ln t)(cos t) between the points ((2+(2pi)^1/2))^2)/4 , 0) and (4 , 0) Solve for the length of this curve and then use this information to help you set up the integral to compute the surface area of this curve when rotated about the y-axis.

 

[BigGlenntheHeavy]

amay9210, if you are desirous of using this message board for succor, not that there is any guarantee that you'll get it, then it behooves you to post your query in a describable nature.
This is incorrect ((2+(2pi)^1/2))^2)/4,0), as the parentheses do not match.

 

[amay9210]

sorry the points that curve is between are ((2+sqarroot 2pi)^2)/4 , 0) and the other point is (4,0)

 

[Deleted member 4993]

Consider the curve given by x(t) = (1 + t^1/2)2 and y(t) = (ln t)(cos t) between the points ((2+(2pi)^1/2))^2)/4 , 0) and (4 , 0) Solve for the length of this curve and then use this information to help you set up the integral to compute the surface area of this curve when rotated about the y-axis.

\(\displaystyle S = \int^{t_2}_{t_1}\sqrt{\left (\frac{dy}{dt}\right )^2 + \left (\frac{dx}{dt}\right )^2} dt\)

Please show us your work, indicating exactly where you are stuck - so that we know where to begin help you.

 

[amay9210]

i got dx/dt = (1+squarroot(t))/squarroot(t)

and dy/dt= -ln(t)sin(t)+(cos(t)/(t))

so ds= squarroot( (dx/dt)^2+(dy/dt)^2)
ds=squarroot ((1+squarroot(t))/squarroot(t))^2 + (-ln(t)sin(t)+(cos(t)/(t)))^2

so length=integral of ds
= integral squarroot ((1+squarroot(t))/squarroot(t))^2 + (-ln(t)sin(t)+(cos(t)/(t)))^2

i don't know how to find the bounds or evaluate this integral through....

 

[BigGlenntheHeavy]

\(\displaystyle amay9210, \ you \ have \ ((2+sqarroot \ 2pi)^2)/4,0), \ this \ is \ still \ incorrect.\)

\(\displaystyle Parentheses \ do \ not \ match.\)