Question

Determine the area to the left of g(y) = 3-y2 and to the right of x = -1

Answer 1

$$-1=3-y^2$$

$$y_1=-2, y_2=2$$

$$A=\displaystyle\int_{-2}^{2}(3-y^2-(-1))dy=[4y-\dfrac{y^3}{3}]\begin{matrix} 2 \\ -2 \end{matrix}$$

$$=8-\dfrac{2^3}{3}-(-8+\dfrac{2^3}{3})=\dfrac{32}{3}({units}^2)$$