Question

Find the area included between the circle $x^{2} + y^{2} = 8x$, parabola $y^{2} = 4x$ and upper part of $x$-axis.

Hint:

For area bounded by two curves, always take $\int (x_{\text{right}} - x_{\text{left}}) dy$.

Answer 1

Step 1: Rewrite the equations. \[ \text{Circle: } x^{2} + y^{2} = 8x $\Rightarrow$ (x-4)^{2} + y^{2} = 16 \] This is a circle with centre $(4,0)$ and radius $4$. \[ \text{Parabola: } y^{2} = 4x $\Rightarrow$ x = \frac{y^{2}}{4} \]

Step 2: Points of intersection. Substitute $x = \frac{y^{2}}{4}$ in circle equation: \[ \left(\frac{y^{2}}{4}\right)^{2} + y^{2} = 8\left(\frac{y^{2}}{4}\right) \] \[ \frac{y^{4}}{16} + y^{2} = 2y^{2} $\Rightarrow$ \frac{y^{4}}{16} = y^{2} \] \[ y^{2}\left(\frac{y^{2}}{16} - 1\right) = 0 $\Rightarrow$ y = 0, \; y = \pm 4 \] On $x$-axis, $y=0 $\Rightarrow$ x=0$. For $y=4$, $x = \frac{16}{4} = 4$. So, points of intersection: $(0,0)$ and $(4,4)$.

Step 3: Required area. We want area bounded by parabola, circle, and $x$-axis above. In terms of $y$: \[ \text{Parabola: } x = \frac{y^{2}}{4}, \text{Circle: } x = 4 + \sqrt{16 - y^{2}} \] Thus, area is: \[ A = \int_{0}^{4} \left[ \big(4 + \sqrt{16 - y^{2}}\big) - \frac{y^{2}}{4} \right] dy \]

Step 4: Solve integration. \[ A = \int_{0}^{4} \left(4 - \frac{y^{2}}{4}\right) dy + \int_{0}^{4} \sqrt{16 - y^{2}} \, dy \] First part: \[ \int_{0}^{4} \left(4 - \frac{y^{2}}{4}\right) dy = \left[4y - \frac{y^{3}}{12}\right]_{0}^{4} = 16 - \frac{64}{12} = 16 - \frac{16}{3} = \frac{32}{3} \] Second part: \[ \int_{0}^{4} \sqrt{16 - y^{2}} \, dy \] This represents a quarter circle of radius $4$, so area = $\frac{1}{4} \pi r^{2} = 4\pi$. Total Area: \[ A = \frac{32}{3} + 4\pi \]

Final Answer: \[ \boxed{\; \dfrac{32}{3} + 4\pi \;} \]