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DryRun
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Homework Statement
Evaluate \int3x^2dx+2xydy, where C is the curve x^2+y^2=4 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:
y=+\sqrt{4-x^2} since the quarter-circle is above the x-axis.
\frac{dy}{dx}=-\frac{x}{\sqrt{4-x^2}}
Replacing y and dy in the line integral.
\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}
Is this correct?
Evaluate \int3x^2dx+2xydy, where C is the curve x^2+y^2=4 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:
y=+\sqrt{4-x^2} since the quarter-circle is above the x-axis.
\frac{dy}{dx}=-\frac{x}{\sqrt{4-x^2}}
Replacing y and dy in the line integral.
\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}
Is this correct?
