Question

Let $\alpha$ be the area of the larger region bounded by the curve $y^2=8 x$ and the lines $y=x$ and $x=2$, which lies in the first quadrant Then the value of $3 \alpha$ is equal to _____

Answer 1

Correct Answer: 22

To solve this problem, we seek the area of the larger region bounded by the parabola $y^2=8x$, the line $y=x$, and the vertical line $x=2$ in the first quadrant. We will calculate this area by integrating the difference between the parabola and the line.

First, find points of intersection: $y^2=8x$ and $y=x$. Substituting, $x^2=8x$, so $x(x-8)=0$. Hence, $x=0$ or $x=8$. In the first quadrant, points are $(0,0)$ and $(8,8)$. The line $x=2$ is vertical.

Now, focus on region from $x=2$ to $x=8$. To find the area under the curve $y^2=8x$ or $y=\sqrt{8x}$, integrate from $x=2$ to $x=8$: $$ A_1 = \int_{2}^{8}\sqrt{8x}\,dx. $$ Use substitution: let $u=8x$, $du=8\,dx$, so $dx=\frac{du}{8}$. The limits of $u$ are $16$ to $64$. Thus, $$ A_1 = \frac{1}{8} \int_{16}^{64} \sqrt{u} \, du = \frac{1}{8} \left[ \frac{2}{3} u^{\frac{3}{2}} \right]_{16}^{64}. $$ Evaluate it: $$ \frac{1}{8} \left( \frac{2}{3} \cdot 64^{\frac{3}{2}} - \frac{2}{3} \cdot 16^{\frac{3}{2}} \right) = \frac{1}{8} \left( \frac{2}{3} \cdot 512 - \frac{2}{3} \cdot 64 \right) = \frac{1}{8} \left( \frac{1024}{3} - \frac{128}{3} \right) = \frac{896}{24} = \frac{112}{3}. $$

Now find the area under $y=x$ from $x=2$ to $x=8$: $$A_2 = \int_{2}^{8} x \, dx = \left[ \frac{x^2}{2} \right]_{2}^{8} = \frac{64}{2} - \frac{4}{2} = 32 - 2 = 30. $$

The area of the region is $A = A_1 - A_2 = \frac{112}{3} - 30 = \frac{112 - 90}{3} = \frac{22}{3}$. Thus, $3\alpha = 3 \times \frac{22}{3} = 22$.

Therefore, $3\alpha$ equals 22, confirming it falls within the given range (22,22).