Question

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

Answer 1

The area is `72` square units.

Explanation:

We have:

`A = int_0^6 xsqrt(36 - x^2)dx`

Now let `u = 36 - x^2`. Then `du = -2xdx` and `dx= (du)/(-2x)`.

`A = int_36^0 xsqrt(u) * (du)/(-2x)`

`A = 1/2int_0^36sqrt(u) du`

`A = 1/2[2/3u^(3/2)]_0^36`

`A = 1/2[2/3(36- x^2)^(3/2)]_0^6`

`A = (1/2)2/3(36 - 0)^(3/2)`

`A = (1/2)144`

`A = 72`

Hopefully this helps!