Can someone please help me with this Calc 3 problem?

∫(x=-√2 to 0)∫(y=-√(4-x²) to 0)√(x²+y²+5)dydx

The region of integration is:http://www.wolframalpha.com/input/?i=y%E2%89%A5-%E...

∫(x=0 to 2)∫(y=-√(4-x²) to 0)√(x²+y²+5)dydx

Region of integration:http://www.wolframalpha.com/input/?i=y%E2%89%A5-%E...

Converting to polar coordinates:

∫(θ=5π/4 to 3π/2)∫(r=0 to 2)√(r²+5) r drdθ + ∫(θ=3π/2 to 2π)∫(r=0 to 2)√(r²+5) r drdθ

= ∫(θ=5π/4 to 2π)∫(r=0 to 2)√(r²+5) r drdθ

u=r²+5

du=2rdr

= 1/2 ∫(θ=5π/4 to 2π)∫(r=0 to 2)√(r²+5) 2r drdθ

= 1/2 ∫(θ=5π/4 to 2π)∫(u=5 to 9)√u dudθ

= 1/2 ∫(θ=5π/4 to 2π) (2/3) u^(3/2) [5 9] dθ

= 1/3 ∫(θ=5π/4 to 2π) 9^(3/2) - 5^(3/2) dθ

= 1/3 ∫(θ=5π/4 to 2π) 27 - 5^(3/2) dθ

= 1/3 (27 - 5^(3/2)) ∫(θ=5π/4 to 2π)dθ

= 1/3 (27 - 5^(3/2)) (2π - 5π/4)

= 1/3 (27 - 5^(3/2)) (3π/4)

= (27 - 5^(3/2))(π/4)

≈ 12.424732.

x = r cos t

y = r sin t

dx dy = r dr dt

x^2 + y^2 = r^2

what is going on with these limits of integration?

y = x

y = -sqrt (4-x^2)

x = -sqrt 2

x = 0

this is 1/8 of a circle or radius 2 from -3pi/4 to -pi/2 (or 5pi/4 to 3pi/2)

we can substitute and solve in polar coordinates if this is not obvious

r sin t = r cos t

tan t = 1

t = pi/4, 5pi / 4

r sin t = -sqrt(4-r cos^2 t)

r^2 sin^2 t = 4 - r^2 cos^2 t

r^2 = 4

r = 2

since tan t >0 and sin t< 0 t is in Q III.

and the other integral

y=0

y = -sqrt (4-x^2)

x = 0

x = 2

this is the quarter circle from -pi/2 to 0 (or 3pi/2 to 2 pi)

∫ ∫ (r^2 + 5 )^1/2 r dr dt

r = 0 to 2

t = -3pi/4 to 0

∫ (1/3)(r^2 + 5)^(3/2) dt

∫ (1/3)(9)^(3/2) dt

∫ 9 dt

(3/4 pi)(9)

27/4 pi