4.2 find the area bounded by the parabola $x = 8 + 2y - y^{2}$, the $y$-axis, and the lines $y=-1$ and $y =…

Explanation:

Step1: Recall the area formula

The area (A) between a curve (x = f(y)), the (y)-axis and the lines (y = a) and (y = b) is given by (A=\int_{a}^{b}|f(y)|dy). Here (f(y)=8 + 2y-y^{2}), (a=-1) and (b = 3).

Step2: Set up the integral

We set up the definite - integral (A=\int_{-1}^{3}(8 + 2y-y^{2})dy).

Step3: Integrate term - by - term

Using the power rule (\int y^{n}dy=\frac{y^{n + 1}}{n+1}+C) ((n\neq - 1)), we have: (\int(8 + 2y-y^{2})dy=8y+y^{2}-\frac{y^{3}}{3}+C).

Step4: Evaluate the definite integral

[ \begin{align*} \left[8y+y^{2}-\frac{y^{3}}{3}\right]_{-1}^{3}&=\left(8\times3+3^{2}-\frac{3^{3}}{3}\right)-\left(8\times(-1)+(-1)^{2}-\frac{(-1)^{3}}{3}\right)\ &=(24 + 9-9)-(-8 + 1+\frac{1}{3})\ &=24-(-7+\frac{1}{3})\ &=24-(-\frac{21 - 1}{3})\ &=24+\frac{20}{3}\ &=\frac{72 + 20}{3}\ &=\frac{92}{3} \end{align*} ]

Answer:

(\frac{92}{3})