Question

Find the area bounded by the parabola x = 8 + 2y − y²; the y-axis and the lines y = −1 and y = 3.

Answer 1

The parabola cuts y-axis at (0, 4) and (0, −2).
Also, the points of intersection of the parabola and the lines y = 3 and y = −1 are B(5, 3) and D(5, −1) respectively.
Therefore, the area of the required region ABCDE
\[A = \int_{- 1}^3 x d y\]
\[ = \int_{- 1}^3 \left( 8 + 2y - y^2 \right) d y\]
\[ = \left[ 8y + y^2 - \frac{y^3}{3} \right]_{- 1}^3 \]
\[ = \left\{ 8\left( 3 \right) + \left( 3 \right)^2 - \frac{\left( 3 \right)^3}{3} \right\} - \left\{ 8\left( - 1 \right) + \left( - 1 \right)^2 - \frac{\left( - 1 \right)^3}{3} \right\}\]
\[ = \left\{ 24 + 9 - 9 \right\} - \left\{ - 8 + 1 + \frac{1}{3} \right\}\]
\[ = \left( 24 \right) - \left\{ - 7 + \frac{1}{3} \right\}\]
\[ = 24 + 7 - \frac{1}{3}\]
\[ = 31 - \frac{1}{3}\]
\[ = \frac{92}{3}\text{ sq . units }\]