Question

Show that `int_0^h int_0^x sqrt(x^2+y^2) dy dx = h^3/6 (sqrt(2) + ln( sqrt(2) + 1) )`?

Answer 1

`int_0^h int_0^x sqrt(x^2+y^2) dy dx = h^3/6 (sqrt(2) + ln( sqrt(2) + 1) )` QED

Explanation:

We want to evaluate:
`int int _R sqrt(x^2+y^2) dA = int_0^h int_0^x sqrt(x^2+y^2) dy dx`

so here `R` is the region:

in the y-direction between `y=0` and `y=x`
in the x-direction between `x=0` and `x=h` (a constant)

This represents a right angled triangle in the first quarter.

If we convert to Polar Coordinates then the region `R` is:

an angle from `theta=0` to `theta=pi/4`
a ray from `r=0` to `r=hsec theta` (as `cos theta = "adj"/"hyp"=h/r`)

And as we convert to Polar coordinates we get:

`x=rcos theta`
`y=rsin theta`
`dA = dy dy = r dr d theta`

So then the integrand `sqrt(x^2+y^2)=sqrt(r^2)=r`

Hence,
`int_0^h int_0^x sqrt(x^2+y^2) dy dx = int_0^(pi/4) int_0^(hsec theta) r r dr d theta`
`" " = int_0^(pi/4) int_0^(hsec theta) r^2 dr d theta`
`" " = int_0^(pi/4) [r^3/3 ]_0^(hsec theta) d theta`
`" " = int_0^(pi/4) ((h sec theta)^3/3 -0) d theta`
`" " = h^3/3 int_0^(pi/4) (sec^3 theta) d theta`
`" " = h^3/3 [1/2sec theta tan theta + 1/2ln|sec theta + tan theta |]_0^(pi/4)`
`" " = h^3/3 {(1/2sec (pi/4) tan (pi/4) + 1/2ln|sec (pi/4) + tan (pi/4) |) - (1/2sec 0 tan 0 + 1/2ln|sec 0 + tan 0 |) }`
`" " = h^3/3 {(1/2*sqrt(2) *1 + 1/2ln| sqrt(2) + 1 |) - (1/2*1*0 + 1/2ln|1 + 0 |) }`
`" " = h^3/3 (1/2*sqrt(2) + 1/2ln( sqrt(2) + 1) )`
`" " = h^3/6 (sqrt(2) + ln( sqrt(2) + 1) )` QED

NOTE
As this is an exercise in performing the double integral, I have used (without proof) the results:

`int sec^3x dx = 1/2secx tanx + 1/2ln|secx + tanx|`

And for clarity

`sec(pi/4)=sqrt(2), tan(pi/4)=1, sec 0=1, tan 0 =0`