I used the equation cos (phi)*sin(phi)*rho d(rho)d(phi)d(theta) with the given boundaries.
I got -632*pi*sqrt(3)/12
But that answer is wrong.
Mackenzie
answeredI misspelled it on the first message.
Steve
answered
∫∫∫ f(p,θ,φ) dv
=∫[0,π/3]∫[0,2π]∫[3,7] p^2 sinφ cosφ dp dθ dφ
= ∫[0,π/3]∫[0,2π] 316/3 * 1/2 sin2φ dθ dφ
= 316π/3 ∫[0,π/3] sin2φ dφ
= -158Ï€/3 (-1/2 - 1)
= 316Ï€
better double-check my math :-)
Steve
answered
Clown Bot
answered
First, let's make sure your integral setup isn't the cause of any monkey business. It seems you have the correct integrand, f(rho, theta, phi) = cos(phi), but let's go over the limits of integration once more.
For rho, you stated 3 < rho < 7, that's spot-on - no confusion there, so kudos to you!
Next, for theta, you have 0 < theta < 2pi. That's also correct, and you're in perfect clown harmony so far.
Here's where things start to get fishy, my friend. For phi, you claimed 0 < phi < pi/3. However, the correct limits for phi should be 0 < phi < pi/3, because pi/3 is actually the upper bound.
Now, let's update the limits and recalculate, shall we?
The integral setup is:
∫∫∫ f(rho, theta, phi) * rho^2 * sin(phi) d(rho) d(phi) d(theta),
with the new limits of integration being:
3 < rho < 7, 0 < theta < 2pi, 0 < phi < pi/3.
Alright, let's give this another whirl and see if we can bring a smile to your face with the correct answer! Happy calculating!
Explain Bot
answered
∫∫∫ cos(phi) * rho^2 * sin(phi) d(rho) d(phi) d(theta)
The limits of integration are:
- For rho: 3 to 7
- For theta: 0 to 2Ï€
- For phi: 0 to π/3
Now let's evaluate the integral step by step:
∫∫∫ cos(phi) * rho^2 * sin(phi) d(rho) d(phi) d(theta)
Step 1: Evaluate the innermost integral with respect to rho (keeping phi and theta constant):
∫ cos(phi) * rho^2 * sin(phi) d(rho)
= (1/3) * cos(phi) * rho^3 * sin(phi) + C
Step 2: Integrate the result from step 1 with respect to phi (keeping theta constant):
∫ (1/3) * cos(phi) * rho^3 * sin(phi) + C d(phi)
= (1/3) * sin(phi) * cos(phi) * rho^3 + C
Step 3: Integrate the result from step 2 with respect to theta (keeping phi constant):
∫ (1/3) * sin(phi) * cos(phi) * rho^3 + C d(theta)
= (1/3) * sin(phi) * cos(phi) * rho^3 * theta + C
Step 4: Apply the limits of integration for theta: 0 to 2Ï€
∫∫ (1/3) * sin(phi) * cos(phi) * rho^3 * theta + C d(theta)
= (1/3) * sin(phi) * cos(phi) * rho^3 * [theta] (from 0 to 2Ï€)
= (1/3) * sin(phi) * cos(phi) * rho^3 * (2Ï€ - 0)
= (2Ï€/3) * sin(phi) * cos(phi) * rho^3
Step 5: Integrate the result from step 4 with respect to phi:
∫ (2π/3) * sin(phi) * cos(phi) * rho^3 d(phi)
= (2Ï€/3) * [- (1/2) * cos^2(phi)] * rho^3 + C
= - (Ï€/3) * cos^2(phi) * rho^3 + C
Step 6: Apply the limits of integration for phi: 0 to π/3
∫∫ - (π/3) * cos^2(phi) * rho^3 + C d(phi)
= - (Ï€/3) * [- (1/3) * cos^3(phi)] * rho^3 + C
= (Ï€/9) * cos^3(phi) * rho^3 + C
Step 7: Integrate the result from step 6 with respect to rho:
∫ (π/9) * cos^3(phi) * rho^3 + C d(rho)
= (Ï€/9) * (1/4) * cos^3(phi) * rho^4 + C
= (Ï€/36) * cos^3(phi) * rho^4 + C
Step 8: Apply the limits of integration for rho: 3 to 7
∫∫∫ (π/36) * cos^3(phi) * rho^4 + C d(rho) d(phi) d(theta)
= (Ï€/36) * (1/5) * cos^3(phi) * rho^5 + C
= (Ï€/180) * cos^3(phi) * rho^5 + C
Now we can substitute the limits of integration:
(Ï€/180) * cos^3(phi) * rho^5 | (rho=7, phi=Ï€/3) - (Ï€/180) * cos^3(phi) * rho^5 | (rho=3, phi=Ï€/3)
Plugging in the values and simplifying, we get:
(Ï€/180) * (cos^3(Ï€/3) * 7^5 - cos^3(Ï€/3) * 3^5)
= (Ï€/180) * [(1/2)^3 * 7^5 - (1/2)^3 * 3^5]
= (Ï€/180) * (1/8) * (7^5 - 3^5)
= (Ï€/1440) * (16807 - 243)
= (Ï€/1440) * 16564
= 362Ï€/720
= π/2
Therefore, the correct answer to the triple integral is π/2, not -632π√3/12.
