Question

Find the area of the region bounded by x² = 16y, y = 1, y = 4 and the y-axis in the first quadrant.

 

Answer 1

\[x^2 = 16 y\text{ is a parabola, with vertex at O}\left( 0, 0 \right)\text{ and symmetrical about + ve }y -\text{ axis }\]
\[y =\text{ 1 is line parallel to }x -\text{ axis cutting the parabola at }\left( - 4, 1 \right)\text{ and }\left( 4, 1 \right)\]
\[ y = 4\text{ is line parallel to }x \text{ axis cutting the parabola at }\left( - 8, 1 \right)\text{ and }\left( 8, 1 \right)\]
\[\text{ Consider a horizontal strip of length }= \left| x \right| \text{ and width }= dy\]
\[ \therefore\text{ Area of approximating rectangle }= \left| x \right| dy\]
\[\text{ The approximating rectangle moves from }y = 1\text{ to } y = 4\]
\[\text{ Area of the curve in the first quadrant enclosed by }y = 1\text{ and }y = 4\text{ is the shaded area }\]
\[ \therefore \text{ Area of the shaded region }= \int_1^4 \left| x \right| dy\]
\[ \Rightarrow A = \int_1^4 x dy ...............\left[ As, x > 0, \left| x \right| = x \right]\]
\[ \Rightarrow A = \int_1^4 \sqrt{16 y} dy\]
\[ \Rightarrow A = 4 \int_1^4 \sqrt{y} dy\]
\[ \Rightarrow A = 4 \left[ \frac{y^\frac{3}{2}}{\frac{3}{2}} \right]_1^4 \]
\[ \Rightarrow A = \frac{8}{3}\left[ 4^\frac{3}{2} - 1^\frac{3}{2} \right]\]
\[ \Rightarrow A = \frac{8}{3} \times 7 = \frac{56}{3}\text{ sq . units }\]
\[ \therefore\text{ Area enclosed by parabola in the first quadrant and }y = 1\text{ and }y = 4\text{ is }\frac{56}{3}\text{ sq . units }\]