Let $${\tmmathbf\vec{F} = yx^2 \tmmathbf\hat{i} + \sin (\pi y) \tmmathbf\hat{j}}$$, and let $${C}$$ be the curve along the line segment starting at (0,2) and ending at (1,4).(adsbygoogle = window.adsbygoogle || []).push({});

$${\int_C \tmmathbf\vec{F} \cdot d \tmmathbf\vec{r} =}$$

My path comes out to be:

r=ti +2tj

dr=i+2j;

F=2t^3i+sin(2*pi*t)j

Now F.dr=2(t^3)+2(sin(2*pi*t))

so

∫2(t^3)+2(sin(2*pi*t))dt=(2t^4 - cos(2*pi*t)/pi ) where t goes from 0 to 1

Note: r,dr and F are vectors.

The answer from the above computation comes out to be 1/2.

I've checked the answer and it's wrong so I've made a mistake somewhere. Could someone identify it?

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# Trouble with line integral

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