Express the volume of the solid that the cylinder r = 4cos(theta) cuts out of the sphere of radius 4 centered at the origin with a triple integral in cylindrical coordinates. I have already found the intervals, but I cannot solve it. The bounds are -sqrt(16-r^2) < z < sqrt(16-r^2), 0 < r < 4cos(theta), -pi/2 < theta < pi/2. Thank you.
Using symmetry, you can just take 4 times the volume in the 1st octant:
4∫[0,π/2]∫[0,4cosθ]∫[0,√(16-r^2)] r dz dr dθ
= 4∫[0,π/2]∫[0,4cosθ] r√(16-r^2) dr dθ
= -4/3∫[0,π/2] (16-r^2)^(3/2) dθ [0,4cosθ]
= -4/3∫[0,π/2] (64sin^3θ)-(64) dθ
= -256/3∫[0,π/2] sin^3θ-1 dθ
= 256/3 ∫[0,π/2] 1- sinθ(1-cos^2θ) dθ
= 256/3 (θ + cosθ - 1/3 cos^3θ) [0,π/2]
= 256/9 (3θ + 3cosθ - cos^3θ) [0,π/2]
= 256/9 ((3π/2)-(3-1))
= 128/9 (3π-4)
To express the volume of the solid, we can set up a triple integral using cylindrical coordinates. The volume element in cylindrical coordinates is given by:
dV = r dz dr dθ
To find the bounds for the triple integral, you have correctly identified the intervals. The bounds you mentioned are:
-√(16 - r²) < z < √(16 - r²)
0 < r < 4cos(θ)
-π/2 < θ < π/2
Using these bounds, we can set up the triple integral as follows:
∫∫∫ r dz dr dθ
Since the function r is given by r = 4cos(θ), we can substitute this expression into the triple integral:
∫∫∫ (4cos(θ)) dz dr dθ
Now, let's evaluate the triple integral using these bounds.
First, integrate with respect to z:
∫∫ (4cos(θ))(√(16 - r²) - (-√(16 - r²))) dr dθ
Simplifying the limits:
∫∫ 8cos(θ)√(16 - r²) dr dθ
Next, integrate with respect to r:
∫ 8cos(θ) ∫√(16 - r²) dr dθ
To integrate with respect to r, we can rewrite as:
∫ 8cos(θ) ∫(16 - r²)^(1/2) dr dθ
Using the power rule for integration:
∫ 8cos(θ) [(16 - r²)^(3/2) / (3/2)] dθ
Simplifying:
∫ 16/3 cos(θ) (16 - r²)^(3/2) dθ
Finally, integrate with respect to θ:
(16/3) ∫ cos(θ) (16 - r²)^(3/2) dθ
Integrating cos(θ) with respect to θ:
(16/3) sin(θ) (16 - r²)^(3/2)
Now, substitute the limits of integration:
(16/3) [sin(π/2) (16 - r²)^(3/2) - sin(-π/2) (16 - r²)^(3/2)]
Since sin(π/2) = 1 and sin(-π/2) = -1, we have:
(16/3) [(16 - r²)^(3/2) + (16 - r²)^(3/2)]
Simplifying further:
(32/3) (16 - r²)^(3/2)
This is the expression for the volume of the solid cut out by the cylinder r = 4cos(θ) from the sphere of radius 4 centered at the origin using a triple integral in cylindrical coordinates.
To express the volume of the solid using a triple integral in cylindrical coordinates, we can set up the integral as follows:
∭dv = ∫∫∫ r dz dr dθ
where the limits of integration are as you have provided:
-√(16-r^2) < z < √(16-r^2)
0 < r < 4cos(θ)
-π/2 < θ < π/2
Now we will proceed with evaluating the integral step-by-step.
Step 1: Evaluate the integral with respect to z:
∭dv = ∫∫ (2√(16-r^2)) dr dθ
Step 2: Evaluate the integral with respect to r:
∭dv = ∫ [2√(16-r^2)](4cos(θ)) dθ
Step 3: Evaluate the integral with respect to θ:
∭dv = ∫∫∫ 8cos(θ)√(16-r^2) dθ dr
To complete the evaluation, the order of integration can be changed, depending on personal preference or ease of calculations.
Note: The integral above represents the volume of the solid cut out from the sphere by the cylinder defined by r = 4cos(θ), with the given limits of integration. Remember to apply the appropriate limits when performing the integration.