AP DSC School Assistant Math Model Test 2

Concept:The area enclosed by a curve and a vertical line above the x-axis is calculated by integrating the positive y-value of the curve with respect to x over the specified interval.Explanation:The given curve is $y^2(2a - x) = x^3$ and the vertical line is $x = 2a$. We need to find the area above the x-axis.First, express y in terms of x. Since the area is above the x-axis, we take the positive square root:$y = \sqrt{\frac{x^3}{2a-x}}$The limits of integration are from $x=0$ (where $y=0$) to $x=2a$ (the given line).The area is given by the integral:$Area = \int_0^{2a} y\,dx = \int_0^{2a} \sqrt{\frac{x^3}{2a-x}}\,dx$To solve this integral, we use the substitution $x = 2a\sin^2\theta$.Then, $dx = 4a\sin\theta\cos\theta\,d\theta$.The limits of integration change as follows:When $x=0$, $2a\sin^2\theta = 0 ⇒ \sin^2\theta = 0 ⇒ \theta = 0$.When $x=2a$, $2a\sin^2\theta = 2a ⇒ \sin^2\theta = 1 ⇒ \theta = \frac{\pi}{2}$.Now, substitute these into the integral:The term $\sqrt{\frac{x^3}{2a-x}}$ becomes:$\sqrt{\frac{(2a\sin^2\theta)^3}{2a - 2a\sin^2\theta}} = \sqrt{\frac{8a^3\sin^6\theta}{2a(1-\sin^2\theta)}} = \sqrt{\frac{8a^3\sin^6\theta}{2a\cos^2\theta}} = \sqrt{4a^2\frac{\sin^6\theta}{\cos^2\theta}} = 2a\frac{\sin^3\theta}{\cos\theta}$The integral for the area becomes:$Area = \int_0^{\pi/2} \left(2a\frac{\sin^3\theta}{\cos\theta}\right) (4a\sin\theta\cos\theta)\,d\theta$$Area = \int_0^{\pi/2} 8a^2\sin^4\theta\,d\theta$$Area = 8a^2 \int_0^{\pi/2} \sin^4\theta\,d\theta$We use the reduction formula for $\int_0^{\pi/2} \sin^n\theta\,d\theta$: $\int_0^{\pi/2} \sin^n\theta\,d\theta = \frac{(n-1)!!}{n!!} \cdot K$, where $K = \frac{\pi}{2}$ if $n$ is even, and $K=1$ if $n$ is odd.For $n=4$: $\int_0^{\pi/2} \sin^4\theta\,d\theta = \frac{(4-1)!!}{4!!} \cdot \frac{\pi}{2} = \frac{3!!}{4!!} \cdot \frac{\pi}{2} = \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} = \frac{3}{8} \cdot \frac{\pi}{2} = \frac{3\pi}{16}$.Substituting this value back into the area calculation:$Area = 8a^2 \cdot \frac{3\pi}{16} = \frac{24\pi a^2}{16} = \frac{3\pi a^2}{2}$Answer:The area enclosed is $\frac{3\pi a^2}{2}$. This corresponds to Option B.