How do you solve this double integral
Favorites|Homepage
Subscriptions | sitemap
HOME > Mathematics > How do you solve this double integral

How do you solve this double integral

[From: ] [author: ] [Date: 11-05-17] [Hit: ]
x^2 + y^2 = 8x ==> r^2 = 8r cos θ ==> r = 8 cos θ.So,= (32/3) (16 - 3π).I hope this helps!......
How do you solve the double integral if f(x,y) = x
and the region you are integrating it over is the region in the first quadrant that lies between x^2+y^2 = 64 and x^2+y^2 = 8x

i know you want to change this to polar coordinates, the double integral is gonna be r^2cos(Θ), and iknow that the bounds for dΘ is gonna be 0 to π/2, but i can't figure out what the bounds are going to be for dr? Can somebody walk me through how to solve this double integral?

-
Note that converting the equations of the circles to polar coordinates yields
x^2 + y^2 = 64 ==> r = 8.
x^2 + y^2 = 8x ==> r^2 = 8r cos θ ==> r = 8 cos θ.

So, ∫∫ x dA
= ∫(θ = 0 to π/2) ∫(r = 8 cos θ to 8) (r cos θ) * (r dr dθ)
= ∫(θ = 0 to π/2) (1/3) r^3 cos θ {for r = 8 cos θ to 8} dθ
= ∫(θ = 0 to π/2) (512/3) (1 - cos^3(θ)) cos θ dθ
= (512/3) ∫(θ = 0 to π/2) [cos θ - cos^4(θ)] dθ
= (512/3) ∫(θ = 0 to π/2) [cos θ - ((1/2)(1 + cos(2θ))^2] dθ
= (512/3) ∫(θ = 0 to π/2) [cos θ - (1/4)(1 + 2 cos(2θ) + cos^2(2θ))] dθ
= (128/3) ∫(θ = 0 to π/2) [4 cos θ - (1 + 2 cos(2θ) + (1/2)(1 + cos(4θ)))] dθ
= (64/3) ∫(θ = 0 to π/2) [8 cos θ - (2 + 4 cos(2θ) + (1 + cos(4θ)))] dθ
= (64/3) ∫(θ = 0 to π/2) [8 cos θ - 3 - 4 cos(2θ) - cos(4θ)] dθ
= (64/3) [8 sin θ - 3θ - 2 sin(2θ) - sin(4θ)/4] {for θ = 0 to π/2}
= (64/3) (8 - 3π/2)
= (32/3) (16 - 3π).

I hope this helps!
1
keywords: solve,you,double,How,this,integral,do,How do you solve this double integral
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .