DryRun
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Homework Statement
Evaluate [itex]\int3x^2dx+2xydy[/itex], where C is the curve [itex]x^2+y^2=4[/itex] starting at (2,0) and ending at (0,2) in the anti-clockwise direction.
The attempt at a solution
The curve C is a quarter-circle with centre (0,0) and radius=2.
Making y subject of formula:
[tex]y=+\sqrt{4-x^2}[/tex] since the quarter-circle is above the x-axis.
[tex]\frac{dy}{dx}=-\frac{x}{\sqrt{4-x^2}}[/tex]
Replacing y and dy in the line integral.
[tex]\int3x^2dx+2xydy=\int_{x=2}^{x=0} 3x^2dx+2x.\sqrt{4-x^2}.-\frac{x}{\sqrt{4-x^2}}.dx=\int_{x=2}^{x=0} x^2dx=\frac{x^3}{3}\Biggr|_2^0=(0-\frac{8}{3})=-\frac{8}{3}[/tex]
Is this correct?
Evaluate [itex]\int3x^2dx+2xydy[/itex], where C is the curve [itex]x^2+y^2=4[/itex] starting at (2,0) and ending at (0,2) in the anti-clockwise direction.
The attempt at a solution
The curve C is a quarter-circle with centre (0,0) and radius=2.
Making y subject of formula:
[tex]y=+\sqrt{4-x^2}[/tex] since the quarter-circle is above the x-axis.
[tex]\frac{dy}{dx}=-\frac{x}{\sqrt{4-x^2}}[/tex]
Replacing y and dy in the line integral.
[tex]\int3x^2dx+2xydy=\int_{x=2}^{x=0} 3x^2dx+2x.\sqrt{4-x^2}.-\frac{x}{\sqrt{4-x^2}}.dx=\int_{x=2}^{x=0} x^2dx=\frac{x^3}{3}\Biggr|_2^0=(0-\frac{8}{3})=-\frac{8}{3}[/tex]
Is this correct?
