Michael Corral: Vector Calculus

4.6 Gradient, Divergence, Curl and Laplacian 185
Step 5: Substitute the formulas for i, j, k from Step 2 and the formulas for ∂F
∂x , ∂F
∂y , ∂F
∂z
from Step 4 into the Cartesian gradient formula∇F(x,y,z)= ∂F
∂x i+∂F ∂y j+∂F ∂z k.
Doing this last step is perhaps the most tedious, since it involves simplifying 3×3+
3×3+2×2=22 terms! Namely,
∇F=
+
1
ρsinφ
(
ρsin 2 φ cosθ ∂F
∂ρ −sinθ∂F +sinφ cosφ cosθ∂F
∂θ ∂φ
+cosφ cosθe φ )
1
ρsinφ
(
ρsin 2 φ sinθ ∂F
∂ρ +cosθ∂F +sinφ cosφ sinθ∂F
∂θ ∂φ
+cosφ sinθe φ )
+ 1 (
ρcosφ ∂F )
ρ ∂ρ −sinφ∂F (cosφe ρ −sinφe φ ),
∂φ
)
(sinφ cosθe ρ −sinθe θ
)
(sinφ sinθe ρ +cosθe θ
which we see has 8 terms involving e ρ , 6 terms involving e θ , and 8 terms involving e φ .
But the algebra is straightforward and yields the desired result:
∇F= ∂F
∂ρ e 1 ∂F
ρ+
ρsinφ ∂θ e θ+ 1 ∂F
ρ∂φ e φ ̌
Example 4.19. In Example 4.17 we showed that∇‖r‖ 2 = 2r and∆‖r‖ 2 = 6, where
r(x,y,z)= xi+yj+zk in Cartesian coordinates. Verify that we get the same answers
if we switch to spherical coordinates.
Solution: Since‖r‖ 2 = x 2 + y 2 + z 2 =ρ 2 in spherical coordinates, let F(ρ,θ,φ)=ρ 2 (so
that F(ρ,θ,φ)=‖r‖ 2 ). The gradient of F in spherical coordinates is
∇F= ∂F
∂ρ e 1 ∂F
ρ+
ρsinφ ∂θ e θ+ 1 ∂F
ρ∂φ e φ
1
= 2ρe ρ +
ρsinφ (0)e θ+ 1 ρ (0)e φ
= 2ρe ρ = 2ρ r , as we showed earlier, so
‖r‖
= 2ρ r = 2r, as expected. And the Laplacian is
ρ
∆F= 1 (
∂
ρ 2 ρ 2∂F
)
1 ∂ 2 (
F 1 ∂
+
∂ρ ∂ρ ρ 2 sin 2 φ ∂θ 2+ ρ 2 sinφ ∂F )
sinφ∂φ
∂φ
= 1 ρ 2 ∂
∂ρ (ρ2 2ρ)+
= 1 ρ 2 ∂
∂ρ (2ρ3 )+0+0
1
ρ 2 sinφ (0)+ 1
ρ 2 sinφ
= 1 ρ 2 (6ρ2 )=6, as expected.
∂
∂φ (sinφ(0))