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
$$\begin{gathered} A={\int }_{-1}^{3}x\mathit{d}y \\ ={\int }_{-1}^3\left(8+2y-y^2\right)\mathit{d}y \\ ={\left[8y+y^2-\frac{y^3}{3}\right]}_{-1}^3 \\ =\left\{8\left(3\right)+{\left(3\right)}^2-\frac{{\left(3\right)}^3}{3}\right\}-\left\{8\left(-1\right)+{\left(-1\right)}^2-\frac{{\left(-1\right)}^3}{3}\right\} \\ =\left\{24+9-9\right\}-\left\{-8+1+\frac{1}{3}\right\} \\ =\left(24\right)-\left\{-7+\frac{1}{3}\right\} \\ =24+7-\frac{1}{3} \\ =31-\frac{1}{3} \\ =\frac{92}{3}\mathrm{sq}.\mathrm{units} \end{gathered}$$