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Calculus 3 Integral!?
If the mass per unit area of a surface is given by ρ=xy, find the mass ∫∫xydS
if S is the part of the cylinder x^2+z^2=16 which is in the first octant and contained within the cylinder x^2+y^2=1.
I m not really sure what bounds to use for this integral.
1 Answer
- kbLv 75 years agoFavorite Answer
∫∫s xy dS
= ∫∫ xy √(1 + (z_x)^2 + (z_y)^2) dA
= ∫∫ xy √(1 + (x/√(16 - x^2))^2 + 0^2) dA, since z = √(16 - x^2)
= ∫∫ xy √(1 + (x^2/(16 - x^2))) dA
= ∫∫ xy √(16/(16 - x^2)) dA
= ∫∫ 4xy dA/√(16 - x^2).
Since the region is bounded by the unit circle x^2 + y^2 = 1 (in the first quadrant), convert to polar coordinates:
∫(r = 0 to 1) ∫(θ = 0 to π/2) 4(r cos θ)(r sin θ) * (r dθ dr)/√(16 - (r cos θ)^2)
= ∫(r = 0 to 1) ∫(θ = 0 to π/2) 2r(2r^2 cos θ sin θ) dθ dr/√(16 - r^2 cos^2(θ))
= ∫(r = 0 to 1) ∫(w = 16-r^2 to 16) 2r dw dr/√w, letting w = 16 - r^2 cos^2(θ)
= ∫(r = 0 to 1) 4rw^(1/2) {for w = 16-r^2 to 16} dr
= ∫(r = 0 to 1) 4r [4 - (16 - r^2)^(1/2)] dr
= ∫(r = 0 to 1) [16r - 4r (16 - r^2)^(1/2)] dr
= [8r^2 + (4/3)(16 - r^2)^(3/2)] {for r = 0 to 1}
= (1/3) [4 * 15^(3/2) - 232]
= (1/3) [4 * 15√15 - 232]
= 20√15 - 232/3.
(Double checked on Wolfram Alpha.)
I hope this helps!
