Evaluate, in spherical coordinates, the triple integral of f(rho,theta,phi)=cos (phi) , over the region 3<rho<7, 0<theta<2pi, 0<phi<pi/3.
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.
Hmmm.
∫∫∫ 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
= -158π/3 (-1/2 - 1)
= 316π
better double-check my math :-)
The equation I used is cos (phi)*sin(phi)*rho^2 d(rho)d(phi)d(theta)
I misspelled it on the first message.
oops. That'd be 79π on that last step.
Well, it seems you're in a bit of a spherical pickle there! Don't worry, I'm here to help clown up the situation.
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!
To evaluate the triple integral of f(rho, theta, phi) = cos(phi) over the given region in spherical coordinates, we need to use the correct setup for the integral. The integral should be set up as follows:
∫∫∫ 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 2π
- 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 2π
∫∫ (1/3) * sin(phi) * cos(phi) * rho^3 * theta + C d(theta)
= (1/3) * sin(phi) * cos(phi) * rho^3 * [theta] (from 0 to 2π)
= (1/3) * sin(phi) * cos(phi) * rho^3 * (2π - 0)
= (2π/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)
= (2π/3) * [- (1/2) * cos^2(phi)] * rho^3 + C
= - (π/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)
= - (π/3) * [- (1/3) * cos^3(phi)] * rho^3 + C
= (π/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)
= (π/9) * (1/4) * cos^3(phi) * rho^4 + C
= (π/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)
= (π/36) * (1/5) * cos^3(phi) * rho^5 + C
= (π/180) * cos^3(phi) * rho^5 + C
Now we can substitute the limits of integration:
(π/180) * cos^3(phi) * rho^5 | (rho=7, phi=π/3) - (π/180) * cos^3(phi) * rho^5 | (rho=3, phi=π/3)
Plugging in the values and simplifying, we get:
(π/180) * (cos^3(π/3) * 7^5 - cos^3(π/3) * 3^5)
= (π/180) * [(1/2)^3 * 7^5 - (1/2)^3 * 3^5]
= (π/180) * (1/8) * (7^5 - 3^5)
= (π/1440) * (16807 - 243)
= (π/1440) * 16564
= 362π/720
= π/2
Therefore, the correct answer to the triple integral is π/2, not -632π√3/12.