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Evaluate the iterated integral by changing to cylindrical coordinates.∫ ^2_0 ∫ ^√(4 − y^2)_0 ∫ ^(16 − x^2 − y^2)_0 1 dz dx dy

User Niran
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To convert the integral to cylindrical coordinates, we use the following conversions:

x = r cos(theta)

y = r sin(theta)

z = z

And we also replace dV with r dz dr d(theta).

The limits of integration are:

0 ≤ r ≤ 2 (since the bounds on x and y are from 0 to 2)

0 ≤ theta ≤ 2pi (since we integrate over the entire circle)

0 ≤ z ≤ 16 - r^2 (since the bounds on z are from 0 to 16 - x^2 - y^2, which in cylindrical coordinates is 16 - r^2)

Thus, the integral becomes:

∫^(2pi)_0 ∫^2_0 ∫^(16-r^2)_0 r dz dr d(theta)

Integrating with respect to z, we get:

∫^(2pi)_0 ∫^2_0 (16 - r^2)r dr d(theta)

Integrating with respect to r, we get:

∫^(2pi)_0 [8r^2 - (1/3)r^4]∣_0^2 d(theta)

= ∫^(2pi)_0 (32/3) d(theta)

= (32/3) ∫^(2pi)_0 d(theta)

= (32/3)(2pi)

= (64/3)pi

Therefore, the value of the iterated integral in cylindrical coordinates is (64/3)pi.

User Derek Liang
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