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Part IA Michaelmas, Lent, Easter Term

Multiple Integrals

Double Integrals

The double integral over a region RR in the xyxy-plane:

∬Rf(x,y) dx dy\iint_R f(x, y)\,dx\,dy

Iterated integrals: For rectangular region a≤x≤ba \leq x \leq b, c≤y≤dc \leq y \leq d:

∬Rf dx dy=∫ab[∫cdf(x,y) dy]dx\iint_R f\,dx\,dy = \int_a^b \left[\int_c^d f(x, y)\,dy\right]dx

Order can often be swapped (Fubini’s theorem).


Triple Integrals

∭Vf(x,y,z) dx dy dz\iiint_V f(x, y, z)\,dx\,dy\,dz

The triple integral gives a volume when f=1f = 1.


Change of Variables

Under transformation (x,y)→(u,v)(x, y) \to (u, v):

∬Rf(x,y) dx dy=∬R′f(x(u,v),y(u,v)) ∣J∣ du dv\iint_R f(x, y)\,dx\,dy = \iint_{R'} f(x(u,v), y(u,v))\,|J|\,du\,dv

where the Jacobian is:

Change of variables showing how area element transforms: dx dy becomes |J| du dv

J=∂(x,y)∂(u,v)=∣∂x∂u∂x∂v∂y∂u∂y∂v∣J = \frac{\partial(x, y)}{\partial(u, v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}

The absolute value ∣J∣|J| accounts for the change in area element.


Polar Coordinates

Transformation: x=rcos⁡θx = r\cos\theta, y=rsin⁡θy = r\sin\theta

Jacobian:

J=∣cos⁡θ−rsin⁡θsin⁡θrcos⁡θ∣=rcos⁡2θ+rsin⁡2θ=rJ = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = r\cos^2\theta + r\sin^2\theta = r

Area element: dA=r dr dθdA = r\,dr\,d\theta

Double integral:

∬Rf dA=∫θ1θ2∫r1(θ)r2(θ)f(rcos⁡θ,rsin⁡θ) r dr dθ\iint_R f\,dA = \int_{\theta_1}^{\theta_2} \int_{r_1(\theta)}^{r_2(\theta)} f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta


Cylindrical Coordinates

(x,y,z)→(r,θ,z)(x, y, z) \to (r, \theta, z) with x=rcos⁡θx = r\cos\theta, y=rsin⁡θy = r\sin\theta, z=zz = z.

Volume element: dV=r dr dθ dzdV = r\,dr\,d\theta\,dz

Useful for: cylinders, circular symmetry about zz-axis.


Spherical Coordinates

(x,y,z)→(r,θ,ϕ)(x, y, z) \to (r, \theta, \phi) with:

  • x=rsin⁡θcos⁡ϕx = r\sin\theta\cos\phi
  • y=rsin⁡θsin⁡ϕy = r\sin\theta\sin\phi
  • z=rcos⁡θz = r\cos\theta

where r≥0r \geq 0, θ∈[0,π]\theta \in [0, \pi], ϕ∈[0,2π)\phi \in [0, 2\pi).

Jacobian: J=r2sin⁡θJ = r^2\sin\theta

Volume element: dV=r2sin⁡θ dr dθ dϕdV = r^2\sin\theta\,dr\,d\theta\,d\phi

Derivation:

J=∣sin⁡θcos⁡ϕrcos⁡θcos⁡ϕ−rsin⁡θsin⁡ϕsin⁡θsin⁡ϕrcos⁡θsin⁡ϕrsin⁡θcos⁡ϕcos⁡θ−rsin⁡θ0∣=r2sin⁡θJ = \begin{vmatrix} \sin\theta\cos\phi & r\cos\theta\cos\phi & -r\sin\theta\sin\phi \\ \sin\theta\sin\phi & r\cos\theta\sin\phi & r\sin\theta\cos\phi \\ \cos\theta & -r\sin\theta & 0 \end{vmatrix} = r^2\sin\theta


Example: Volume of a Sphere

Using spherical coordinates:

V=∭dV=∫02π∫0π∫0Rr2sin⁡θ dr dθ dϕV = \iiint dV = \int_0^{2\pi} \int_0^\pi \int_0^R r^2\sin\theta\,dr\,d\theta\,d\phi

=[r33]0R⋅[−cos⁡θ]0π⋅[ϕ]02π= \left[\frac{r^3}{3}\right]_0^R \cdot \left[-\cos\theta\right]_0^\pi \cdot \left[\phi\right]_0^{2\pi}

=R33⋅2⋅2π=4πR33= \frac{R^3}{3} \cdot 2 \cdot 2\pi = \frac{4\pi R^3}{3}


Example: Gaussian Integral

Evaluate I=∫−∞∞e−x2 dxI = \int_{-\infty}^{\infty} e^{-x^2}\,dx.

Method: Square and convert to polar.

I2=∫−∞∞e−x2 dx∫−∞∞e−y2 dy=∬−∞∞e−(x2+y2) dx dyI^2 = \int_{-\infty}^{\infty} e^{-x^2}\,dx \int_{-\infty}^{\infty} e^{-y^2}\,dy = \iint_{-\infty}^{\infty} e^{-(x^2+y^2)}\,dx\,dy

Convert to polar: x2+y2=r2x^2 + y^2 = r^2, dx dy=r dr dθdx\,dy = r\,dr\,d\theta

I2=∫02π∫0∞e−r2r dr dθ=2π[−12e−r2]0∞=πI^2 = \int_0^{2\pi} \int_0^{\infty} e^{-r^2} r\,dr\,d\theta = 2\pi \left[-\frac{1}{2}e^{-r^2}\right]_0^\infty = \pi

Result:

∫−∞∞e−x2 dx=π\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}


Choosing Coordinates

GeometryBest Coordinates
Rectangular boxCartesian
Cylinder, circular baseCylindrical
Sphere, central forceSpherical
EllipseModified polar or elliptical

Match the coordinate system to the domain boundaries for simplest limits.