- #1

DavidAp

- 44

- 0

**∫∫cos(x^2 + y^2)dA, where R is the region that lies above the x-axis within the circle x^2 + y^2 = 9.**

*Answer: .5pi*sin(9)*

**My Work:**

∫(0 ->pi) ∫(0 -> 9) cos(r^2) rdrdθ

u = r^2

du = 2rdr

dr = du/2r

u = r^2

du = 2rdr

dr = du/2r

.5∫(0 ->pi) ∫(0 -> 9) cos(u) dudθ

.5∫(0 ->pi) sin(u)(0 -> 9) dθ

.5∫(0 ->pi) sin(r^2)(0 -> 9) dθ

.5∫(0 ->pi) [sin(9^2) - sin(0)] dθ

.5∫(0 ->pi) sin(81) dθ

.5[sin(81) θ](0 -> pi)

.5pi*sin(81)

However, the answer is suppose to be .5pi*sin(9). Where did I go wrong? Am I not suppose to square r, and if not then doesn't that mean everything I did below the substitution doesn't work?

Thank you for your time and patience. I apologize for my integral notation, I'm not sure of a better way to type it.