# What the area of the figure bounded by the given curve y=(x*(sqrt36-x^2)) (0_< x_< 6)?

May 11, 2018

The area is $72$ square units.

#### Explanation:

We have:

$A = {\int}_{0}^{6} x \sqrt{36 - {x}^{2}} \mathrm{dx}$

Now let $u = 36 - {x}^{2}$. Then $\mathrm{du} = - 2 x \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{du}}{- 2 x}$.

$A = {\int}_{36}^{0} x \sqrt{u} \cdot \frac{\mathrm{du}}{- 2 x}$

$A = \frac{1}{2} {\int}_{0}^{36} \sqrt{u} \mathrm{du}$

$A = \frac{1}{2} {\left[\frac{2}{3} {u}^{\frac{3}{2}}\right]}_{0}^{36}$

$A = \frac{1}{2} {\left[\frac{2}{3} {\left(36 - {x}^{2}\right)}^{\frac{3}{2}}\right]}_{0}^{6}$

$A = \left(\frac{1}{2}\right) \frac{2}{3} {\left(36 - 0\right)}^{\frac{3}{2}}$

$A = \left(\frac{1}{2}\right) 144$

$A = 72$

Hopefully this helps!