Double integral over circular surface
I've decided to finish my education through completing my last exam (I've
been working for 5 years). The exam is in multivariable calculus and I
took the classes 6 years ago so I am very rusty. Will ask a bunch of
questions over the following weeks and I love you all for helping me.
I got this from an exam answer:
$$\iint\limits_S F\cdot dS=\iint\limits_{x^2+y^2\le1}F(x,y,2)\cdot
(0,0,1)dxdy= 2\pi$$
I presume it goes
$$..\iint\limits_{x^2+y^2\le1}F(x,y,2)\cdot
(0,0,1)dxdy=\iint\limits_{x^2+y^2\le1}0x+0y+2dxdy=\iint\limits_{x^2+y^2\le1}2dxdy=..$$
But how to think for that last step to get $2\pi$? I have never solved a
double integral over a joined surface like that. I relise that $2\pi$ is a
full circle, but I would like to know exactly why I get that answer in
this case.
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