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I need to calculate the line integral of (x^2 - x*y)dy when the curve is y=x^2 from (-1,1) to (2,4) If I convert the x's into y ie x^2=y and x=(y)^1/2 and use the limits 1 to 4 I get -4.9

If I use x as a parameter and convert the y into x and replace dy by 2xdx and use limits -1 to 2 I do not get -4.9

Why is this so ?

user379001
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1 Answers1

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$$x=t\;,\;\;y=t^2\;,\;\;-1\le t\le 2\implies \int_C (x^2-xy)dy=\int_{-1}^2(t^2-t^3)2tdt=$$

$$=2\int_{-1}^2(t^3-t^4)dt=\left[\frac12t^4-\frac25t^5\right]_{-1}^2=\frac12(16-1)-\frac25(32-(-1))=$$

$$=\frac{15}2-\frac{66}5=\frac{75-132}{10}=-\frac{57}{10}$$

I don't get $\;-4.9\;$ , either.

DonAntonio
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  • What's wrong with x= (y)^(1/2) parameterization. y=x^2 is equal to x=(y)^(1/3) – user379001 Oct 16 '16 at 06:00
  • Sorry it's y=x^2 is equal to x=(y)^(1/2) – user379001 Oct 16 '16 at 06:02
  • @user379001 No, of course not. The equation $;y=x^2;$ defines a straight- upwards parabola and it is defined for any $;x\in\Bbb R;$, whereas $;x=y^{1/2};$ defines just half a horizontal-rightwards parabola (the upper half) and then it always $;x\ge0;$ . – DonAntonio Oct 16 '16 at 08:19
  • So if I evaluate the Integral with x=y^(1/2) and limits 0 to 4 and sum it to the Integral with x=-y^(1/2) and limits 1 to 0 , then that's correct. The answer it gives is -5.7. Is this correct? – user379001 Oct 16 '16 at 20:45