<p>The area between the parabola \(x=4y-y^2\) and the line \(x=y\) is: [MAU014]</p>
Step-by-Step Solution
Key Concept: Intersections: y=4y-y^2 \to y^2-3y=0 \to y=0,3. Integrate w.r.t. y: \int_0^3(4y-y^2-y)dy = \int_0^3(3y-y^2)dy.
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<p>Intersections: $y=4y-y^2\Rightarrow y^2-3y=0\Rightarrow y(y-3)=0\Rightarrow y=0,3$.</p>
<p>On $[0,3]$: $x=4y-y^2\ge x=y$ (since $4y-y^2\ge y\Leftrightarrow 3y-y^2=y(3-y)\ge0$ ✓).</p>
<p>$$A=\int_0^3[(4y-y^2)-y]\,dy=\int_0^3(3y-y^2)\,dy=\left[\frac{3y^2}{2}-\frac{y^3}{3}\right]_0^3=\frac{27}{2}-9=\frac{9}{2}$$</p>
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Correct Answer: C