How can I evaluate :
$$\int_{C} y \;dx + x^2 \; dy$$
where $C$ is the parabola define by
$$y=4x-x^2 \quad \text{from } \; (4,0) \; \text{ to } \; (1,3).$$
Do I need to parameterize the parabola?
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$\begingroup$ Typically you should indicate the parameterization used...in this case there is a natural one which is to use x as the parameter $\endgroup$– TriatticusMar 16, 2016 at 19:24
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$\begingroup$ Note the orientation of the curve as well. $\endgroup$– ncmathsadistMar 16, 2016 at 19:35
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$\begingroup$ Ok... So, how can I do that? $\endgroup$– hlapointeMar 16, 2016 at 19:39
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2 Answers
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$$\because y = 4x-x^2 , \therefore \frac{dy}{dx} = 4-2x , \therefore dy = (4-2x)dx $$ Hence, $$\int_{C} y \;dx + x^2 \; dy = \int_{4}^{1} 4x-x^2 dx + \int_{4}^{1} x^2 (4-2x)dx$$ $$\implies \int_4^1 4x-x^2+4x^2-2x^3$$ $$\implies \int_4^1 4x+3x^2-2x^3$$ $$\implies (2x^2+x^3-\frac{x^4}{2})_4^1$$ $$\implies (2+1-\frac{1}{2}) - (32+64-128)$$ $$\implies 35-\frac{1}{2}$$ $$\implies \frac{69}{2}$$
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$\begingroup$ Isn't $\int_{4}^{1} x^2 \cdot (4-2x) \; dx$? $\endgroup$ Mar 16, 2016 at 21:48
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$\begingroup$ @hlapointe : thanks for pointing it out, i've corrected my answer now. $\endgroup$– rotaivaMar 16, 2016 at 22:29
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hint:
The points on the parabola are $P=(x,y)=(x,4x-x^2)$ for $4\ge x\ge 1$ and $$ dy=\frac{dy}{dx}dx=(4-2x)dx $$
can you do from this?
answered Mar 16, 2016 at 19:48