Spherical coordinates ?
Let V be the volume of the region D that is bounded below by the xy-plane, above
by the sphere x^2 + y^2 + z^2 = 4, and on the sides by the cylinder x^2 + y^2 = 1.
Express V as an iterated triple integral in spherical coordinates in the order '' d'phi' d'rho' d'theta'
What are the steps that i should follow ? Where does the angle 'phi' enter and exit ?
- kbLv 77 years agoFavorite Answer
Converting to spherical coordinates:
x^2 + y^2 + z^2 = 4 ==> ρ = 2.
x^2 + y^2 = 1 ==> ρ sin φ = 1 ==> φ = arcsin(1/ρ).
These surfaces intersect when 1^2 + z^2 = 4
==> z = √3 (since the region is above the x-axis), which is a plane.
In spherical coordinates, this is given by ρ cos φ = √3 ==> φ = arccos(√3/ρ).
Split the region into two parts (sketching this may be useful):
(i) Spherical cap (with φ in [0, arccos(√3/ρ)]).
(ii) Cylindrical base (with φ in [arccos(√3/ρ), arcsin(1/ρ)]).
(At the base z = 0, we have ρ = 1, since the unit circle is 1 unit away from (0,0,0).)
So the volume ∫∫∫ 1 dV equals
∫(θ = 0 to 2π) ∫(ρ = 0 to 2) ∫(φ = 0 to arccos(√3/ρ)) 1 * (ρ^2 sin dφ dρ dθ)
+ ∫(θ = 0 to 2π) ∫(ρ = 1 to 2) ∫(φ = arccos(√3/ρ) to arcsin(1/ρ)) 1 * (ρ^2 sin dφ dρ dθ).
I hope this helps!
- JasanpahafLv 67 years ago
I think it comes in handy to draw/ sketch what is going on, (Although 3D is hard to visualize, when you do it yourself you get the idea better!).
if you do, you'll have a half a hemisphere (bottom part of the sphere) with a hole drilled through (for these, you gotta make sense out of the words. So try practicing some more problems).
Then you have to parameterize (not sure how you spell it). It basically means finding your intervals of integration as wells as your dv. And for that, it is better to change from the conventional x,y,z to "Spherical Coordinates" or it'll be almost impossible to integrate. Now for spherical coordinates:
φ, θ, p;
p is your sphere's radius
θ is how much is covered (in radians) looking from the top view
φ is how much is covered from (+) z-axis to (-) z-axis (side view)
X = p•sin(φ) • cos(θ)
Y = p•sin(φ) • sin(θ )
z = p•cos(φ)
p^2 = x^2 + y^2 + z^2
dV = p^2 sin(φ) dp dθ dφ;
here is a visual link: http://tutorial.math.lamar.edu/Classes/CalcIII/TIS...
you can find P right away: remember that P^2 = x^2+ y^2 + z^2 = 4; so P = 2 and so your interval is
0<= P <= 2, because we integrate from origin and outwards
for theta and phi you get them from the picture (specially φ). In this case, the cylinder won't change θ, and so
0<= θ <= 2 pi ; because looking from the top we get a complete circle.
φ is the tricky interval. and I wish I have more time for that. hope the link helps!