Question

Find the area, lying above x-axis and included between the circle x² + y² = 8x and the parabola y² = 4x.

Answer 1

The given equations are

\[x^2 + y^2 = 8x \cdots\left( 1 \right)\]

\[y^2 = 4x \cdots\left( 2 \right)\]

Clearly the equation \[x^2 + y^2 = 8x\] is a circle with centre
\[\left( 4, 0 \right)\] and has a radius 4. Also \[y^2 = 4x\]  is a parabola with vertex at origin and the axis along the x-axis opening in the positive direction.

To find the intersecting points of the curves ,we solve both the equation.

\[\therefore\] \[x^2 + 4x = 8x\]

\[\Rightarrow\] \[x^2 - 4x = 0\]

\[\Rightarrow\] \[x\left( x - 4 \right) = 0\]

\[\Rightarrow\] \[x = 0\text{ and }x = 4\]

When \[x = 4, y = \pm 4\] To approximate the area of the shaded region the length \[= \left| y_2 - y_1 \right|\]  and the width = dx

\[A = \int_0^4 \left| y_2 - y_1 \right| d x\]

\[= \int_0^4 \left( y_2 - y_1 \right) dx  ...........\left[ \because y_2 > y_1 \therefore \left| y_2 - y_1 \right| = y_2 - y_1 \right]\]

\[= \int_0^4 \left[ \sqrt{\left( 16 - \left( x - 4 \right)^2 \right)} - \sqrt{4x} \right] d x .............\left\{ \therefore y_2 = \sqrt{16 - \left( x - 4 \right)^2}\text{ and }y_1 = 2\sqrt{x} \right\}\]

\[= \int_0^4 \sqrt{16 - \left( x - 4 \right)^2} d x - \int_0^4 \sqrt{4x} d x\]

\[= \left[ \frac{\left( x - 4 \right)}{2}\sqrt{16 - \left( x - 4 \right)^2} + \frac{16}{2} \sin^{- 1} \left( \frac{x - 4}{4} \right) \right]^4_0 - \left[ \frac{4 x^\frac{3}{2}}{3} \right]^4_0\]

\[= \left[ 0 + 0 - 0 - 8 \sin^{- 1} \left( \frac{- 4}{4} \right) \right] - \frac{4}{3} \times 4^\frac{3}{2}\]

\[= \frac{8\pi}{2} - \frac{32}{3}\]

\[ = 4\pi - \frac{32}{3}\]

Hence the required area is  \[4\pi - \frac{32}{3}\] square units.