Question

R={(x,y)/-1<=x<=1, -2<=y<=2}
∫∫(1-x²)^(1/2)dxdy = ?

Answer 1

trig sub, yeah?

Answer 2

trying to solve without using trig substitution

Answer 3

\[\int\limits \sqrt{a^2 - x^2} dx = \frac{1}{2} x \sqrt{a^2-x^2} +\frac{1}{2}a^2\tan^{-1}\frac{x}{\sqrt{a^2-x^2}}+C\]seems a good formula to work off of

Answer 4

\[\int\limits_{-1}^1\int\limits_{-2}^2 \sqrt{1^2 - x^2} dydx = 4\int\limits_{-1}^1\sqrt{1^2 - x^2} dx\]\[= 2 x \sqrt{1-x^2} +2\tan^{-1}\frac{x}{\sqrt{1-x^2}}\]evaluated from -1 to 1
I don't want to evaluate it, can't you?

Answer 5

Trying to understand what you did there. Where did that 4 come from?

Answer 6

evaluating the y part:
\[\int\limits_{-2}^2\int\limits_{-1}^1 \sqrt{1-x^2}dydx=\int\limits_{-1}^1 y \sqrt{1-x^2}dx\]evaluated from -2 to 2 is\[\int\limits\limits_{-1}^1 2 \sqrt{1-x^2}-(-2) \sqrt{1-x^2}dx=4\int\limits\limits_{-1}^1 \sqrt{1-x^2}dx\]

Answer 7

Oh, stupid me, that's obvious. I was seeing things differently. Thank you.