2 V16-х 4 -y" 3/2
Please help me evaluate this
We are given the integral,
We will be solving the above integral by converting it into cylindrical coordinates.
Note:
For the cylindrical co-ordinate, x = r cos(θ), y = r sin(θ), and z = z.
And dV = dz dy dx = r dz dr dθ.
Here, 0 ≤ x ≤ 2,
0 ≤ y ≤ √(16 - x2),
and 0 ≤ z ≤ 4.
Notice that, 0 ≤ y ≤ √(16 - x2) => 0 ≤ y2 ≤ 16 - x2 (squaring both sides)
=> 0 ≤ x2 + y2 ≤ 16
=> 0 ≤ r2 ≤ 16
=> 0 ≤ r ≤ 4
Further, 0 ≤ x ≤ 2 => 0 ≤ x ≤ 2
=> 0 ≤ r cos(θ) ≤ 2
=> 0 ≤ cos(θ) ≤ 1/2 (As 0 ≤ r ≤ 4)
We know, cos-1(0) = π/2 and cos-1(1/2) = π/3.
=> π/3 ≤ θ ≤ π/2.
Summary:
Take, x = r cos(θ), y = r sin(θ), and z = z.
Then, dz dy dx = r dz dr dθ.
And here, 0 ≤ z ≤ 4, 0 ≤ r ≤ 4, and π/3 ≤ θ ≤ π/2.
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