evaluate the triple integral of
sqrt(x^2+y^2) dV where V is the region that lies inside the cylinder x^2 +y^2 = 16 and between the planes z=-5 and z=4.
I understand that I must use cylindrical coordinates but I keep getting the wrong answer.
Can someone work all the way through it and explain the whole thing please. Thanks!
sqrt(x^2+y^2) dV where V is the region that lies inside the cylinder x^2 +y^2 = 16 and between the planes z=-5 and z=4.
I understand that I must use cylindrical coordinates but I keep getting the wrong answer.
Can someone work all the way through it and explain the whole thing please. Thanks!
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∫∫∫ √(x^2 + y^2) dV x^2 + y^2 = 16 and 4 < z < 5
use cylindrical coordinates
x = r cosθ , y = r sin θ 0 < θ < 2π 0 < r < 4 and 4 < z < 5
∫∫∫ r r dz dr dθ
= ∫∫∫ r^2 dz dr dθ 4 < z < 5, 0 < r < 4 and 0 < θ < 2π
= ∫∫ r^2 z [ from 4 to 5 ] dr dθ
= ∫∫ r^2 ( 5 - 4) dr dθ
= ∫∫ r^2 dr dθ 0 < r < 4 and 0 < θ < 2π
= ∫(1/3) r^3 [ from 0 to 4 ] dθ 0 < θ < 2π
= ∫(1/3) 64dθ 0 < θ < 2π
= (64/3)θ from 0 to 2π
= (64/3) (2π)
= 128π /3
use cylindrical coordinates
x = r cosθ , y = r sin θ 0 < θ < 2π 0 < r < 4 and 4 < z < 5
∫∫∫ r r dz dr dθ
= ∫∫∫ r^2 dz dr dθ 4 < z < 5, 0 < r < 4 and 0 < θ < 2π
= ∫∫ r^2 z [ from 4 to 5 ] dr dθ
= ∫∫ r^2 ( 5 - 4) dr dθ
= ∫∫ r^2 dr dθ 0 < r < 4 and 0 < θ < 2π
= ∫(1/3) r^3 [ from 0 to 4 ] dθ 0 < θ < 2π
= ∫(1/3) 64dθ 0 < θ < 2π
= (64/3)θ from 0 to 2π
= (64/3) (2π)
= 128π /3