Question

Find the area bounded by the curve y = 4 − x² and the lines y = 0, y = 3.

Answer 1

\[y = 4 - x^2\text{ is a parabola, with vertex (0, 4), opening downwars and having axis of symmetry as - ve }y -\text{ axis }\]
\[y = 0\text{ is the }x - \text{ axis, cutting the parabola at }A(2, 0)\text{ and A'}( - 2, 0)\]
\[y = 3\text{ is a line parallel to }x - \text{ axis, cutting the parabola at B }(1, 3 )\text{ and B'}( - 1, 3) \text{ and }y -\text{ axis at }C(0, 3) \]
\[\text{ Required area is the shaded area ABB'A }= 2 \left(\text{ area ABCO }\right)\]
\[\text{ Consider a horizontal strip of length }= \left| x_2 - x_1 \right|\text{ and width = dy in the shaded region }\]
\[\text{ Area of approximating rectangle }= \left| x_2 - x_1 \right| dy\]
\[\text{ The approximating rectangle moves from }y = 0\text{ to }y = 3 \]
\[ \therefore\text{ Area of shaded region }= 2 \int_3^0 \left| x_2 - x_1 \right| dy \]
\[ \Rightarrow A = 2 \int_0^3 \left( x_2 - x_1 \right) dy ...............\left[ As, \left| x_2 - x_1 \right| = x_2 - x_{1 ,} x_2 > x_1 \right] \]
\[ \Rightarrow A = 2 \int_0^3 \left( \sqrt{4 - y} - 0 \right) dy\]
\[ \Rightarrow A = - 2 \left[ \frac{\left( 4 - y \right)^\frac{3}{2}}{\frac{3}{2}} \right]_0^3 \]
\[ \Rightarrow A = - 2 \left[ \frac{\left( 4 - y \right)^\frac{3}{2}}{\frac{3}{2}} \right]_0^3 \]
\[ \Rightarrow A = 2 \times \frac{2}{3}\left[ 4^\frac{3}{2} - 1^\frac{3}{2} \right]\]
\[ \Rightarrow A = \frac{4}{3} \times 7 \]
\[ \Rightarrow A = \frac{28}{3}\text{ sq . units }\]
\[ \therefore \text{ Area bounded by the two parabolas }= \frac{28}{3}\text{ sq . units }\]