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Triple Integration (spherical)

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Spherical coordinates
Wize Concept
Spherical coordinates. A point P=(x,y,z)P = (x, y, z)P=(x,y,z) is represented by the ordered triple (ρ,θ,ϕ)(\rho, \theta, \phi)(ρ,θ,ϕ) where
ρ=distance from P=(x,y,z) to O=(0,0,0)ϕ=angle between positive z−axis and OPθ=polar angle of (x,y) in the xy−plane\begin{array}{l}
\rho = distance\ from\ P = (x, y, z)\ to\ O = (0, 0, 0)\\[5pt]
\phi = angle\ between\ positive\ z-axis\ and\ OP\\[5pt]
\theta = polar\ angle\ of\ (x, y)\ in\ the\ xy-plane
\end{array}ρ=distance from P=(x,y,z) to O=(0,0,0)ϕ=angle between positive z−axis and OPθ=polar angle of (x,y) in the xy−plane​
x=ρsin⁡ϕcos⁡θρ=x2+y2+z2y=ρsin⁡ϕsin⁡θcos⁡θ=xρsin⁡ϕz=ρcos⁡ϕcos⁡ϕ=zρ\begin{array}{ll}
x=\rho\sin\phi\cos\theta & \rho=\sqrt{x^2+y^2+z^2}\\[5pt]
y=\rho\sin\phi\sin\theta & \cos\theta=\dfrac{x}{\rho\sin\phi}\\[7pt]
z=\rho\cos\phi & \cos\phi=\dfrac{z}{\rho}
\end{array}x=ρsinϕcosθy=ρsinϕsinθz=ρcosϕ​ρ=x2+y2+z2​cosθ=ρsinϕx​cosϕ=ρz​​


→\to→ Notes:
Spherical coordinates are useful when there is symmetry about the origin (e.g. spheres).

Triple Integration (spherical)
Wize Concept
Triple Integration (spherical). Let D⊂R3D \subset \mathbb{R}^3D⊂R3 be defined by
g1(θ,ϕ)≤ρ≤g2(θ,ϕ),a≤ϕ≤b,c≤θ≤dg_1(\theta, \phi) \le \rho \le g_2(\theta, \phi),\quad a\le \phi \le b,\quad c \le \theta \le dg1​(θ,ϕ)≤ρ≤g2​(θ,ϕ),a≤ϕ≤b,c≤θ≤d
where gig_igi​ is continuous. If f(x,y,z)f(x, y, z)f(x,y,z) is continuous on DDD, then
∭Df(x,y,z)dV=∫ab∫cd∫g1(θ,ϕ)g2(θ,ϕ)f(ρsin⁡ϕcos⁡θ,ρsin⁡ϕsin⁡θ,ρcos⁡ϕ)ρ2sin⁡ϕdρdϕdθ\iiint\limits_{D}f(x,y,z)dV=\int_a^b\int_c^d\int_{g_1(\theta,\phi)}^{g_2(\theta,\phi)}f(\rho\sin\phi\cos\theta,\rho\sin\phi\sin\theta,\rho\cos\phi)\rho^2\sin\phi d\rho d\phi d\thetaD∭​f(x,y,z)dV=∫ab​∫cd​∫g1​(θ,ϕ)g2​(θ,ϕ)​f(ρsinϕcosθ,ρsinϕsinθ,ρcosϕ)ρ2sinϕdρdϕdθ

→\to→ Notes:
As with iterated integrals, the integral can be rearranged; take care to change the bounds of integration as well.
Don't forget the extra factor of ρ2sin⁡ϕ\rho^2 \sin \phiρ2sinϕ in the integrand.

Practice: Triple integration (spherical)
Represent the solid DDD bounded by z2=x2+y2z^2 = x^2 + y^2z2=x2+y2, x2+y2+z2=1x^2 + y^2 + z^2 = 1x2+y2+z2=1, and z≥0z \ge 0z≥0 using spherical coordinates.

∫θ=0θ=2π∫ϕ=0ϕ=π4∫ρ=0ρ=1ρ2sin⁡ϕ dρ dϕ dθ\displaystyle \int_{\theta=0}^{\theta=2\pi}\int_{\phi=0}^{\phi=\frac{\pi}{4}}\int_{\rho=0}^{\rho=1}\rho^2\sin\phi\ d\rho\ d\phi\ d\theta∫θ=0θ=2π​∫ϕ=0ϕ=4π​​∫ρ=0ρ=1​ρ2sinϕ dρ dϕ dθ

∫θ=0θ=2π∫ϕ=0ϕ=π4∫ρ=0ρ=1ρ2cos⁡ρ dρ dϕ dθ\displaystyle \int_{\theta=0}^{\theta=2\pi}\int_{\phi=0}^{\phi=\frac{\pi}{4}}\int_{\rho=0}^{\rho=1}\rho^2\cos\rho\ d\rho\ d\phi\ d\theta∫θ=0θ=2π​∫ϕ=0ϕ=4π​​∫ρ=0ρ=1​ρ2cosρ dρ dϕ dθ

∫θ=0θ=2π∫ϕ=0ϕ=π4∫ρ=−1ρ=1ρ2cos⁡ρ dρ dϕ dθ\displaystyle \int_{\theta=0}^{\theta=2\pi}\int_{\phi=0}^{\phi=\frac{\pi}{4}}\int_{\rho=-1}^{\rho=1}\rho^2\cos\rho\ d\rho\ d\phi\ d\theta∫θ=0θ=2π​∫ϕ=0ϕ=4π​​∫ρ=−1ρ=1​ρ2cosρ dρ dϕ dθ

∫θ=0θ=2π∫ϕ=0ϕ=π4∫ρ=−1ρ=1ρ2 dρ dϕ dθ\displaystyle \int_{\theta=0}^{\theta=2\pi}\int_{\phi=0}^{\phi=\frac{\pi}{4}}\int_{\rho=-1}^{\rho=1}\rho^2\ d\rho\ d\phi\ d\theta∫θ=0θ=2π​∫ϕ=0ϕ=4π​​∫ρ=−1ρ=1​ρ2 dρ dϕ dθ

None of the above

I don't know

Practice: Triple integration (spherical)
Let DDD be the solid bounded by z2=x2+y2z^2 = x^2 + y^2z2=x2+y2, x2+y2+z2=1x^2 + y^2 + z^2 = 1x2+y2+z2=1, and z≥0z \ge 0z≥0. Evaluate 
∭DzdV\iiint\limits_{D}zdVD∭​zdV

Practice: Triple integration (spherical)
Write down an integral using spherical coordinates representing ∭E(x2+y2)   ⁣dV\displaystyle\iiint_E(x^2+y^2) \ \de{V}∭E​(x2+y2)  dV where EEE is the part of the positive octant inside the sphere x2+y2+z2=4x^2+y^2+z^2 = 4x2+y2+z2=4, outside the sphere x2+y2+z2=2x^2+y^2+z^2 = 2x2+y2+z2=2, and inside the cone 2z=x2+y22z =\sqrt{x^2 + y^2}2z=x2+y2​.

∫0π/2∫0arctan⁡2∫22ρ4sin⁡3ϕ dρ dϕ dθ\displaystyle\int^{\pi/2}_0\int^{\arctan 2}_0\int^2_{\sqrt{2}}\rho^4\sin^3\phi\ d\rho\ d\phi\ d\theta∫0π/2​∫0arctan2​∫2​2​ρ4sin3ϕ dρ dϕ dθ

∫0π∫0arctan⁡2∫22ρ4sin⁡3ϕ dρ dϕ dθ\displaystyle\int^{\pi}_0\int^{\arctan 2}_0\int^2_{\sqrt{2}}\rho^4\sin^3\phi\ d\rho\ d\phi\ d\theta∫0π​∫0arctan2​∫2​2​ρ4sin3ϕ dρ dϕ dθ

∫0π∫0arctan⁡2∫22ρ3sin⁡3ϕ dρ dϕ dθ\displaystyle\int^{\pi}_0\int^{\arctan 2}_0\int^2_{\sqrt{2}}\rho^3\sin^3\phi\ d\rho\ d\phi\ d\theta∫0π​∫0arctan2​∫2​2​ρ3sin3ϕ dρ dϕ dθ

∫0π/2∫0arctan⁡2∫22ρ3sin⁡3ϕ dρ dϕ dθ\displaystyle\int^{\pi/2}_0\int^{\arctan 2}_0\int^2_{\sqrt{2}}\rho^3\sin^3\phi\ d\rho\ d\phi\ d\theta∫0π/2​∫0arctan2​∫2​2​ρ3sin3ϕ dρ dϕ dθ

None of the above

I don't know

Extra Practice

Practice: Triple integration (spherical)

Practice: Triple integration (spherical)
Write down an integral using spherical coordinates representing ∭E(x2+y2)   ⁣dV\displaystyle\iiint_E(x^2+y^2) \ \de{V}∭E​(x2+y2)  dV where EEE is the part of the positive octant inside the sphere x2+y2+z2=4x^2+y^2+z^2 = 4x2+y2+z2=4, outside the sphere x2+y2+z2=2x^2+y^2+z^2 = 2x2+y2+z2=2, and inside the cone 2z=x2+y22z =\sqrt{x^2 + y^2}2z=x2+y2​.

Practice: Triple integration (spherical)

Practice: Triple integration (spherical)
Write down an integral using spherical coordinates representing ∭E(x2+y2)   ⁣dV\displaystyle\iiint_E(x^2+y^2) \ \de{V}∭E​(x2+y2)  dV where EEE is the part of the positive octant inside the sphere x2+y2+z2=4x^2+y^2+z^2 = 4x2+y2+z2=4, outside the sphere x2+y2+z2=2x^2+y^2+z^2 = 2x2+y2+z2=2, and inside the cone 2z=x2+y22z =\sqrt{x^2 + y^2}2z=x2+y2​.

Practice: Triple integration (spherical)

Practice: Triple integration (spherical)
Write down an integral using spherical coordinates representing ∭E(x2+y2)   ⁣dV\displaystyle\iiint_E(x^2+y^2) \ \de{V}∭E​(x2+y2)  dV where EEE is the part of the positive octant inside the sphere x2+y2+z2=4x^2+y^2+z^2 = 4x2+y2+z2=4, outside the sphere x2+y2+z2=2x^2+y^2+z^2 = 2x2+y2+z2=2, and inside the cone 2z=x2+y22z =\sqrt{x^2 + y^2}2z=x2+y2​.