Question

How do you find the definite integral for: `(9-x^2)^(1/2)` for the intervals `[-3, 3]`?

Answer 1

The answer is `=(9pi)/2`

Explanation:

First, we calculate `intsqrt(9-x^2)dx`
let `x=3sinu`, then `dx=3cosudu`
`sqrt(9-9sin^2u)=3cosu`
`:.intsqrt(9-x^2)dx=int3cosu*3cosudu`
`=9intcos^2udu`
`cos2u=2cos^2u-1` `=>` `cos^2u=(cos2u+1)/2`
`9intcos^2udu=9/2int(cos2u+1)du`
`=9/2((sin2u)/2+u)= 9/4((sin2u)+2u)`
`sin2u=2sinucosu=2*x/3*sqrt(9-x^2)/3=2/9x(sqrt(9-x^2))`
`9/4((sin2u)+2u)=9/4*2/9*(xsqrt(9-x^2))+9/2arcsin(x/3)`
`int_-3^3sqrt(9-x^2)dx=(1/2(xsqrt(9-x^2))+9/2arcsin(x/3))_-3^3`
`0+9/2*arcsin1-0-9/2arcsin(-1)`
`0+9pi/4--9pi/4=9pi/2`

Answer 2

`(9pi)/2`

Explanation:

We can also think about this geometrically. `y=sqrt(9-x^2)` is the upper half of the circle `x^2+y^2=9`, a circle with radius `3` and center at `(0,0)`.

Since the bounds span the length of the semicircle, the integral will be equal to `1//2` the circle's area, or `(9pi)/2`.