Question

1) Find the area bounded by the curve y=9-x and the x axis.

a) Horizontal Strip

b) Vertical Strip

Answer 1

1)

$$9-x^2=0=>x_1=-3, x_2=3$$

a)

$$y=9-x^2=>x=-\sqrt{9-y}\ or \ x=\sqrt{9-y}$$ $$A=\displaystyle\int_{0}^{9}(\sqrt{9-y}-(-\sqrt{9-y}))dy$$ $$=2\displaystyle\int_{0}^{9}\sqrt{9-y}dy=2\big[-\dfrac{2}{3}(9-y)^{3/2}\big]\begin{matrix} 9 \\ 0 \end{matrix}$$ $$=-\dfrac{4}{3}((9-9)^{3/2}-(9-0)^{3/2})=36({units}^2)$$

b)

$$A=\displaystyle\int_{-3}^{3}(9-x^2)dx=\big[9x-\dfrac{x^3}{3}\big]\begin{matrix} 3 \\ -3 \end{matrix}$$ $$9(3)-\dfrac{(3)^3}{3}-(9(-3)-\dfrac{(-3)^3}{3})=36({units}^2)$$