Find the volume of the solid from the graph of the equation below.
$$2\sqrt {xy} $$
and above the square with the vertices (1,1) (1,2) (2,2) (2,1)
$$A = \{(x,y): 1 \le x \le 2 , 1 \le y \le 2\}$$ On double integrating the above equation we get the integral as below, I've chosen the strips/arrow parallel to x-axis.
$$\int_1^2\int_1^2(2\sqrt{xy}) dxdy $$
Could anyone please tell me if the integral has been set up correctly?
[NOTE: I'm supposed to use double integrals only]
$\int_1^2\int_1^2(2\sqrt{xy}) dxdy\\ \int_1^2\int_1^2(2\sqrt{x}\sqrt {y}) dxdy$
I am breaking up the radical because when we integrate with respect to $x, y$ will be treated as a constant.
$ \int_1^2(2 (\frac23)(x)^\frac 32\sqrt {y})|_1^2dy\\ \int_1^2 (\frac43)(2^\frac 32 - 1)\sqrt {y} \ dy\\ $
And now you have your old friend single variable calculus.
Since your region for y does not depend on your region for $x,$ it would be permisslbe to do this.
$2(\int_1^2\sqrt y\ dy)(\int_1^2\sqrt{x}\ dx)\\ 2(\int_1^2\sqrt{x}\ dx)^2 $
But until you understand why that works, I don't suggest you get in the habbit of doing that.