Wize University Calculus 3 Textbook > Multiple Integrals

Cylindrical coordinates in three dimensional space is similar to the polar coordinates in two dimensional space:
Cylindrical coordinates are related to Cartesian coordinates as follows:
x=rcos⁡θy=rsin⁡θz=z\begin{array}{c}
x=r\cos \theta\\[10pt]
y=r\sin\theta\\[10pt]
z=z
\end{array}x=rcosθy=rsinθz=z​
Similar to polar coordinates:
r2=x2+y2dV=dxdydz=dr⋅rdθ⋅dz\begin{array}{c}
r^2=x^2+y^2\\[10pt]
dV=dxdydz=dr\cdot rd\theta\cdot dz
\end{array}r2=x2+y2dV=dxdydz=dr⋅rdθ⋅dz​
The following integration could be rewritten in terms of cylindrical coordinates:
∭Ef(x,y,z)dV=∬D[∫z1(z,y)z2(x,y)f(x,y,z)dz]dA\iiint_Ef(x,y,z)dV=\iint_D\bigg[\int_{z_1(z,y)}^{z_2(x,y)}f(x,y,z)dz\bigg]dA∭E​f(x,y,z)dV=∬D​[∫z1​(z,y)z2​(x,y)​f(x,y,z)dz]dA
if region DDD is defined as:
D={(r,θ)∣α≤θ≤β, g1(θ)≤r≤g2(θ)}D = \{(r, \theta)|\alpha \le \theta \le \beta,\ g_1(\theta) \le r \le g_2(\theta) \}D={(r,θ)∣α≤θ≤β, g1​(θ)≤r≤g2​(θ)}
Then, the general form of a triple integral using cylindrical coordinates will be:
∭Ef(x,y,z)dV=∫αβ∫g1(θ)g2(θ)∫z1(rcos⁡θ,rsin⁡θ)z2(rcos⁡θ,rsin⁡θ)f(rcos⁡θ,rsin⁡θ,z)rdrdθdz\iiint_Ef(x,y,z)dV=\int_\alpha^\beta\int_{g_1(\theta)}^{g_2(\theta)}\int_{z_1(r\cos\theta,r\sin\theta)}^{z_2(r\cos\theta,r\sin\theta)}f(r\cos\theta,r\sin\theta,z)rdrd\theta dz∭E​f(x,y,z)dV=∫αβ​∫g1​(θ)g2​(θ)​∫z1​(rcosθ,rsinθ)z2​(rcosθ,rsinθ)​f(rcosθ,rsinθ,z)rdrdθdz

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A sketch of spherical space is shown below:
Spherical coordinates are related to Cartesian coordinates as follows:
x=ρsin⁡ϕcos⁡θy=ρsin⁡ϕsin⁡θz=ρcos⁡ϕ\begin{array}{c}
x=\rho\sin\phi\cos\theta\\[10pt]
y=\rho\sin\phi\sin\theta\\[10pt]
z=\rho\cos\phi
\end{array}x=ρsinϕcosθy=ρsinϕsinθz=ρcosϕ​
where that θ\thetaθ is the angle in the x−yx-yx−y space.
ϕ\phiϕ is the angle measured from the positive zzz-axis.
θ\thetaθ is the distance from the origin.
We also have:
ρ2=x2+y2+z2dV=ρ2sin⁡ϕdϕdρdθ=ρsin⁡ϕdθ⋅ρdϕ⋅dρ\begin{array}{c}
\rho^2=x^2+y^2+z^2\\[10pt]
dV=\rho^2\sin\phi d\phi d\rho d\theta=\rho\sin\phi d\theta\cdot\rho d\phi\cdot d\rho
\end{array}ρ2=x2+y2+z2dV=ρ2sinϕdϕdρdθ=ρsinϕdθ⋅ρdϕ⋅dρ​

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If region EEE in spherical space is defined as follows:
E={(ρ,θ,ϕ)∣a≤rho≤b, α≤θ≤β, λ≤ϕ≤μ}E=\{(\rho,\theta,\phi)|a\le rho\le b,\ \alpha\le\theta\le\beta,\ \lambda\le\phi\le\mu\}E={(ρ,θ,ϕ)∣a≤rho≤b, α≤θ≤β, λ≤ϕ≤μ}
Then, the triple integral of f(x,y,z)f(x,y,z)f(x,y,z) over EEE is equal to:
∭Ef(x,y,z)dV=∫λμ∫αβ∫abf(ρsin⁡ϕcos⁡θ,ρsin⁡ϕsin⁡θ,ρcos⁡ϕ)ρ2sin⁡ϕdρdθdϕ\iiint_Ef(x,y,z)dV=\int_\lambda^\mu\int_\alpha^\beta\int_a^bf(\rho\sin\phi\cos\theta,\rho\sin\phi\sin\theta,\rho\cos\phi)\rho^2\sin\phi d\rho d\theta d\phi∭E​f(x,y,z)dV=∫λμ​∫αβ​∫ab​f(ρsinϕcosθ,ρsinϕsinθ,ρcosϕ)ρ2sinϕdρdθdϕ

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Calculate the volume under z=x2+y2z=\sqrt{x^2+y^2}z=x2+y2​ within the cylinder x2+y2=4x^2+y^2=4x2+y2=4, and between the planes z=0z=0z=0 and z=4z=4z=4.

x2+y2=r2=4⇒0≤r≤2x^2+y^2=r^2=4\Rightarrow 0\leq r\leq 2x2+y2=r2=4⇒0≤r≤2

x2+y2=r2=r\sqrt{x^2+y^2}=\sqrt{r^2}=rx2+y2​=r2​=r

V=∫02πdθ∫02dr∫0r  r  dz\displaystyle V=\int_0^{2\pi} d\theta\int_0^2 dr\int_0^r\;r\;dzV=∫02π​dθ∫02​dr∫0r​rdz=∫02πdθ∫02dr  rz∣0r\displaystyle=    \int_0^{2\pi} d\theta\int_0^2 dr\;rz\bigg\vert_0^r=∫02π​dθ∫02​drrz​0r​

=∫02πdθ∫02dr  r2=∫02πdθ(r3/3)∣02\displaystyle=\int_0^{2\pi} d\theta\int_0^2 dr\;r^2=    \int_0^{2\pi} d\theta(r^3/3)\bigg\vert_0^2=∫02π​dθ∫02​drr2=∫02π​dθ(r3/3)​02​=∫02πdθ8/3=83∫02πdθ=16π3\displaystyle=\int_0^{2\pi} d\theta8/3=\frac{8}{3}\int_0^{2\pi} d\theta=\frac{16\pi}{3}=∫02π​dθ8/3=38​∫02π​dθ=316π​

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Calculate the volume of the solid that is above the cone z=x2+y2z=\sqrt{x^2+y^2}z=x2+y2​ and below the sphere x2+y2+z2=1x^2+y^2+z^2=1x2+y2+z2=1.

0≤θ≤2π0\le\theta\le2\pi0≤θ≤2π
0≤ρ≤10\le\rho\le10≤ρ≤1
x2+y2+z2=ρ2=1x^2+y^2+z^2=\rho^2=1x2+y2+z2=ρ2=1
z≥x2+y2⇒ρcos⁡(ϕ)≥ρsin⁡(ϕ)z\geq\sqrt{x^2+y^2}\Rightarrow \rho\cos(\phi)\geq\rho\sin(\phi)z≥x2+y2​⇒ρcos(ϕ)≥ρsin(ϕ)
cos⁡(ϕ)≥sin⁡(ϕ)⇒tan⁡(ϕ)≤1⇒0≤ϕ≤π/4\cos(\phi)\geq\sin(\phi)\Rightarrow \tan(\phi)\leq1\Rightarrow 0\leq\phi\leq\pi/4cos(ϕ)≥sin(ϕ)⇒tan(ϕ)≤1⇒0≤ϕ≤π/4

V=∭EdV=∫02π  dθ∫0π/4  dϕ∫01  dρ  ρ2sin⁡(ϕ)\displaystyle V=\iiint_E dV=\int_0^{2\pi}\;d\theta\int_0^{\pi/4}\;d\phi\int_0^{1}\;d\rho\;\rho^2\sin(\phi)V=∭E​dV=∫02π​dθ∫0π/4​dϕ∫01​dρρ2sin(ϕ)
=∫02π  dθ∫0π/4sin⁡(ϕ)  dϕ∫01  dρ  ρ2=2π(−cos⁡(ϕ))∣0π/4⋅13ρ3∣01\displaystyle=\int_0^{2\pi}\;d\theta\int_0^{\pi/4}\sin(\phi)\;d\phi\int_0^{1}\;d\rho\;\rho^2=    2\pi(-\cos(\phi))\bigg\vert_0^{\pi/4}\cdot\frac{1}{3}\rho^3\bigg\vert_0^1=∫02π​dθ∫0π/4​sin(ϕ)dϕ∫01​dρρ2=2π(−cos(ϕ))​0π/4​⋅31​ρ3​01​
=2π[−cos⁡(π/4)+cos⁡(0)]⋅13(1)3=2π3[1−22]=2\pi[-\cos(\pi/4)+\cos(0)]\cdot\frac{1}{3}(1)^3=\frac{2\pi}{3}[1-\frac{\sqrt{2}}{2}]=2π[−cos(π/4)+cos(0)]⋅31​(1)3=32π​[1−22​​]

Evaluate ∫05∫−5−x20∫x2+y2−119−3x2−3y2(2x−3y)dzdydx\displaystyle\int_{0}^{\sqrt{5}}\int_{-\sqrt{5-x^2}}^{0}\int_{x^2+y^2-11}^{9-3x^2-3y^2}(2x-3y)dzdydx∫05​​∫−5−x2​0​∫x2+y2−119−3x2−3y2​(2x−3y)dzdydxusing cylindrical coordinates.

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Evaluate ∫∫∫x2+y2dVE                            \displaystyle\int\int\int x^2+y^2dV\newline{}
\scriptsize{E}~~~~~~~~~~~~~~~~~~~~~~~~~~~~∫∫∫x2+y2dVE                            , where E is the region portion at x2+y2+z2=4x^2+y^2+z^2=4x2+y2+z2=4 with y≥0.y\geq0.y≥0.

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Let S be the region on the first octant which lies above z2=x2+y2z^2=x^2+y^2z2=x2+y2 and below (z−1)2+x2+y2=1.(z-1)^2+x^2+y^2=1.(z−1)2+x2+y2=1.Express in:

a) Cylindrical Coordinates
b) Spherical Coordinates
c) Evaluate using either

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