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

Question

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

1 Answer

Impact of this question

1141 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-11T16:41:04

The area is $72$ $72$ square units.

Explanation:

We have:

$A = \int_{0}^{6} x \sqrt{36 - x^{2}} d x$ $A = int_0^6 xsqrt(36 - x^2)dx$

Now let $u = 36 - x^{2}$ $u = 36 - x^2$ . Then $d u = - 2 x d x$ $du = -2xdx$ and $d x = \frac{d u}{- 2 x}$ $dx= (du)/(-2x)$ .

$A = \int_{36}^{0} x \sqrt{u} \cdot \frac{d u}{- 2 x}$ $A = int_36^0 xsqrt(u) * (du)/(-2x)$

$A = \frac{1}{2} \int_{0}^{36} \sqrt{u} d u$ $A = 1/2int_0^36sqrt(u) du$

$A = \frac{1}{2} {[\frac{2}{3} u^{\frac{3}{2}}]}_{0}^{\frac{3}{2}}$ $A = 1/2[2/3u^(3/2)]_0^36$

$A = \frac{1}{2} {[\frac{2}{3} {(36 - x^{2})}^{\frac{3}{2}}]}_{0}^{\frac{3}{2}}$ $A = 1/2[2/3(36- x^2)^(3/2)]_0^6$

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

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

$A = 72$ $A = 72$

Hopefully this helps!

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question