# Homework Help: How to integrate int ydx + zdy over a helix?

1. Feb 3, 2008

### i2837856393

1. The problem statement, all variables and given/known data

$$\int$$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.

2. Feb 3, 2008

### 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?

Last edited: Feb 3, 2008
3. Feb 3, 2008

### i2837856393

i'm sorry you're right...
that should be pi/2 - pi which is -pi/2