Question

Double Integral Of (sqrt[sqrt(x)+sqrt(y)]dxdy = ???
Domain of Integral is: R= sqrt(x) + sqrt(y) <= 1
plzzzz help!

Answer 1

use equation of circle and you'll have the limits of x as 0 and y= sqrt (x^2+1) and limit of Y as 0 to 1

Answer 2

\[\int\limits_{0}^{1}\int\limits_{0}^{\sqrt{1-x^2}} \sqrt{1-x^2}dxdy \]

Answer 3

can you do it now?

Answer 4

no wait

Answer 5

\[\int \int \sqrt{\sqrt{x}+\sqrt{y}} \ \text{d}x \text{d}y\]

Answer 6

\[({\sqrt x+ \sqrt y})^{1/2}\]

Answer 7

take y as constant whilst integrating with respect to x first

Answer 8

mapping with \(x=u^2\) and \(y=v^2\) makes the domain \(D :u+v \le 1\) and\[\int \int_R \sqrt{\sqrt{x}+\sqrt{y}} \ \text{d}x \text{d}y=\int \int_D \left| \frac{\partial(x,y)}{\partial(u,v)} \right|\sqrt{u+v} \ \text{d}u \text{d}v\]

Answer 9

i think this is better for integrating now\[\int_{0}^{1} \int_{0}^{1-v} 4uv\sqrt{u+v} \ \text{d}u \text{d}v\]

Answer 10

I got nasty value
int(int(sqrt(sqrt(x)+sqrt(y)), x = 1 .. (1-sqrt(y))^2), y = 0 .. 1)

(64/315)*x^(9/4)+(64/105)*x^(7/4)-(64/315)*x^(3/2)*(sqrt(x)+1)^(3/2)-(32/105)*x*(sqrt(x)+1)^(3/2)+(8/15)*sqrt(x)*(sqrt(x)+1)^(3/2)-(44/63)*(sqrt(x)+1)^(3/2)

Answer 11

what is it?

Answer 12

i guess the method stated above ..that is by method of conversion would be easier

Answer 13

yeah, i think i solved it. i convert x = u^4 and y=v^4. then turned it to polar mode.

Answer 14

no need to go for polar mode

Answer 15

i just figured it out with x = u^4 and y=v^4.this is much better.

Answer 16

that int 0^1 got messed up

Answer 17

~~~test~~~\[\large\oint_C f(x,y)\,\text{d}s\]\[\large\iint\limits_{S}\!\!\!\!\!\!\!\!\!\!\!\!\subset\!\!\supset f(x,y,z)\,\text{d}A\]\[\large\int\!\!\!\!\int\!\!\!\!\int\limits_{\!\!\!\!\!\! \partial V}f(x,y,z,t)\,\text{d}V\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\huge{\subset\!\!\supset}\]\[\large\int\!\!\!\!\int\!\!\!\!\int\!\!\!\!\int \limits_{\!\!\!\!\!\!\!\!\!\!\! \partial H}f(x,y,z,t,\psi)\,\text{d}H\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\huge{\subset \;\supset}\]