How to integrate int ydx + zdy over a helix?

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SUMMARY

The integral of the form ∫ydx + zdy over the helix defined by r(t) = (sin t)i + (cos t)j + tk, for 0 ≤ t ≤ π, was evaluated incorrectly by a forum user. The correct evaluation of the integral leads to the result of 3π, while the user initially computed it as 3(π/2). The confusion arose from miscalculating the limits of integration, specifically the evaluation of the definite integral which should yield π/2 - π, resulting in -π/2. The discussion highlights the importance of careful limit application in integral calculus.

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Homework Statement



\intydx + 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.
 
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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:
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.
 

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