Setting up polar coordinate double integrals
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Setting up polar coordinate double integrals

[From: ] [author: ] [Date: 12-11-01] [Hit: ]
2 ∫(θ = 0 to π/2) ∫(r = 0 to 4 cos θ) r * (r dr dθ),= 2 ∫(u = 0 to 1) (64/3) (1 - u^2) du,= 256/9.I hope this helps!......
I'm having trouble setting up my integrals for polar coordinates. I have a quiz tomorrow on this stuff and a step-by-step explanation to the following would be awesome:
Find the volume V of the region by using iterated integrals in polar coordinates: The solid region above the xy plane bounded on the sides by the cylinder x^2 + y^2 -4x = 0 and above by the cone z^2 = x^2 + y^2
The answer in the back of the book is 256/9.
Thanks in advance guys.

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In polar coordinates:
z^2 = x^2 + y^2 ==> z = r.

x^2 + y^2 - 4x = 0 ==> r^2 - 4r cos θ = 0
==> r = 4 cos θ (which is completely traced out for θ in [0, π]).

So, the volume ∫∫ √(x^2 + y^2) dA equals
2 ∫(θ = 0 to π/2) ∫(r = 0 to 4 cos θ) r * (r dr dθ), via polar coordinates (and symmetry)
= 2 ∫(θ = 0 to π/2) (1/3)r^3 {for r = 0 to 4 cos θ} dθ
= 2 ∫(θ = 0 to π/2) (64/3) cos^3(θ) dθ
= 2 ∫(θ = 0 to π/2) (64/3) (1 - sin^2(θ)) * cos θ dθ
= 2 ∫(u = 0 to 1) (64/3) (1 - u^2) du, letting u = sin θ
= (128/3) (u - u^3/3) {for u = 0 to 1}
= 256/9.

I hope this helps!
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