How to integrate int ydx + zdy over a helix?

[i2837856393]

Homework Statement

$$\int$$ydx + zdy
C

where C is the part of the helix r(t)=(sin t)(i) + (cos t)(j) + t(k)
and 0 < t < pi (those should be greater than or equal to signs)


$$\int$$(cost)(cost)-t(sint)
$$\int$$(1+cos 2t)/2 - t sint
with limits from zero to pi.

this then equaled to
1/2t + (sin 2t)/4 + tcost -sin t


and this gave me the answer 3(pi)/2
but the answer should be 3 pi..
can anyone else solve this and check please or tell me where i went wrong.

 

[Dick]

I agree with your indefinite integral. But if i put the limits in I get -pi/2. That's still not 3*pi...? Is there a typo in the problem?

 

[i2837856393]

i'm sorry you're right...
that should be pi/2 - pi which is -pi/2
i accidentally added them...
yea. then you and i are in agreement. i guess the teacher made a mistake. I've spent hours and this was the only solution i came to.