Question

a curve is defined as x=cos(t)+t*sin(t),y=sin(t)-t*cos(t), FInd the area of surface of revolution if t is between 0 and a half pi, and the curve is rotated around the x-axis

Answer 1

define (x')^2 and (y')^2

Answer 2

dx / dt = t cos t, dy / dt = t sin t
That's what the mark scheme said

Answer 3

^2 them and take the sqrt

Answer 4

should end up with t i think

Answer 5

but the derivatives are not right, are they?

Answer 6

x = cos(t)+t sin(t)
x' = -sin + sin + t cos

y = sin(t)-t cos(t)
y' = cos - cos + t sin

i dont see why they wouldnt be

Answer 7

t^2(cos^2+sin^2) = t^2

and sqrt(t^2) = t
\[2\pi\int yt~ dt\]

Answer 8

OMG!!!! My mistake ! I just get the derivative wrong! I'm soooooo stupid!

Answer 9

practice helps, and then we still make mistakes at times :)

Answer 10

Now I need to evaluate \[\int\limits_{?}^{?} t ^{2} \cos t\]

Answer 11

How to do it ? Confused

Answer 12

by parts

du int v
------------
cos t
+ t^2 sin t
- 2t -cos t
+ 2 -sin t
- 0

so...

t^2 sin t + 2t cos t - 2 sin t