Double Integrals Over Rectangles

Geometry

We have z=f(x,y) defined on [a,b]×[c,d][a,b] \times [c,d][a,b]×[c,d]. We name the domain to be D, and cut it into smaller rectangles. Take an arbitrary point (xij,yij)(x_{ij}, y_{ij})(xij,yij) on the rectangle [xi,xi+1]×[yj,yj+1][x_i, x_{i+1}] \times [y_j,y_{j+1}][xi,xi+1]×[yj,yj+1]. We calculate the Riemann sum.
∑i=1m∑j=1nf(xij,yij)(xi+1−xi)(yj+1−yj)=∑i=1m∑j=1nf(xij,yij)b−amd−cn\sum_{i=1}^m\sum_{j=1}^n f(x_{ij}, y_{ij}) (x_{i+1} - x_i)(y_{j+1} - y_{j}) \\\\ = \sum_{i=1}^m\sum_{j=1}^n f(x_{ij}, y_{ij}) \frac{b-a}{m} \frac{d-c}{n} \\\\ i=1∑mj=1∑nf(xij,yij)(xi+1−xi)(yj+1−yj)=i=1∑mj=1∑nf(xij,yij)mb−and−c
If such a limit exists, then we define it to be the double integral of f over the rectangle D. This is the volume between f and xy-plane on D. If the value of the function is always 1 in D, then the double integral equals to the area of D.

Iterated Integrals

Introduce a function A(x)=∫cdf(x,y)dyA(x) = \int_c^d f(x,y) dyA(x)=∫cdf(x,y)dy. Next, we integrate A. ∫abA(x)dx=∫ab[∫cdf(x,y)dy]dx\int_a^b A(x) dx = \int_a^b [\int_c^d f(x,y) dy] dx∫abA(x)dx=∫ab[∫cdf(x,y)dy]dx. Alternatively, we can introduce a function of B(y)B(y)B(y) to integrate on x first.

Fubini’s Theorem

Assume f is continuous function on the rectangle D={(x,y)∣a≤x≤b,c≤y≤d}D=\{(x,y)|a\le x\le b, c \le y\le d \}D={(x,y)∣a≤x≤b,c≤y≤d}.
∫∫Df(x,y)dA=∫ab∫cdf(x,y)dydx=∫cd∫abf(x,y)dxdy\int\int_D f(x,y) dA = \int_a^b \int_c^d f(x,y) dydx = \int_c^d \int_a^b f(x,y) dxdy ∫∫Df(x,y)dA=∫ab∫cdf(x,y)dydx=∫cd∫abf(x,y)dxdy
E.x.

Evaluate ∫∫ysin(xy)dA\int \int ysin(xy) dA∫∫ysin(xy)dA where D=[1,2]×[0,π]D = [1,2] \times [0,\pi]D=[1,2]×[0,π].

sol:
∫∫ysin(xy)dA=∫0π∫12ysin(xy)dxdy=∫0π[−cos(xy)]12dy=∫0πcos(y)−cos(2y)dy\int \int ysin(xy) dA \\\\ = \int_0^{\pi} \int_1^2 ysin(xy) dxdy \\\\ = \int_0^{\pi} [-cos(xy)]_1^2 dy \\\\ = \int_0^{\pi} cos(y)-cos(2y) dy \\\\ ∫∫ysin(xy)dA=∫0π∫12ysin(xy)dxdy=∫0π[−cos(xy)]12dy=∫0πcos(y)−cos(2y)dy
The order of doing iterated integral may be crucial.

Double Integrals Over General Regions

We have f(x,y) on D. Assume D is bounded (we can find a rectangle R that encloses D).

Define h(x,y) = f(x,y) if (x,y)∈D(x,y) \in D(x,y)∈D, and h(x,y) = 0 if (x,y)∈(R−D)(x,y) \in (R-D)(x,y)∈(R−D). In this case,
∫∫Df(x,y)dA=∫∫Rh(x,y)dA\int\int_D f(x,y) dA = \int\int_R h(x,y) dA ∫∫Df(x,y)dA=∫∫Rh(x,y)dA
A plane region D is said to be of type I if it lies between the graphs of two continuous functions of x, say g1(x)g_1(x)g1(x) and g2(x)g_2(x)g2(x).

By applying the Fubini’s theorem,
∫∫Df(x,y)dA=∫ab∫g1(x)g2(x)f(x,y)dydx\int\int_D f(x,y) dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y) dydx ∫∫Df(x,y)dA=∫ab∫g1(x)g2(x)f(x,y)dydx
A plane region D is said to be of type II if it lies between the graphs of two continuous functions of y.

E.x.

Evaluate ∫01∫x1sin(y2)dydx\int_0^1 \int_x^1 sin(y^2) dydx∫01∫x1sin(y2)dydx.

sol:

Trick here is to switch the order of integration. Drawing the region D, we find out that it is both type I and type II. Convert the integral.
∫01∫x1sin(y2)dydx=∫01∫0ysin(y2)dxdy=∫01ysin(y2)dy\int_0^1 \int_x^1 sin(y^2) dydx \\\\ = \int_0^1 \int_0^y sin(y^2) dxdy \\\\ = \int_0^1 ysin(y^2) dy \\\\ ∫01∫x1sin(y2)dydx=∫01∫0ysin(y2)dxdy=∫01ysin(y2)dy

Double Integrals In Polar Coordinates

Suppose f is a continuous function defined on D (Draw a picture of D!). Convert the limit of x and y to r and theta in the polar coordinate as 0≤a≤r≤b0 \le a \le r \le b0≤a≤r≤b, α≤β\alpha \le \betaα≤β. The formula is as below
∫∫Df(x,y)dA=∫αβ∫abf(rcosθ,rsinθ)rdrdθ\int\int_D f(x,y) dA = \int_{\alpha}^{\beta} \int_{a}^{b} f(r cos \theta, r sin \theta)r dr d\theta ∫∫Df(x,y)dA=∫αβ∫abf(rcosθ,rsinθ)rdrdθ
If the formula involves z, then simplify it to the form of z=f(x,y). If the original equation involves terms like z2z^2z2, remember the integral of z=g(x,y)z=\sqrt{g(x,y)}z=g(x,y) only calculate the part where z>0z>0z>0. Based on symmetry, we can get the desired answer by multiplying the integral result by 2.

E.x.

Evaluate ∫∫D3x+4y2dA\int \int_D 3x+4y^2 dA∫∫D3x+4y2dA where the region is the upper half plane bounded by the circles x2+y2=1x^2+y^2=1x2+y2=1 and x2+y2=4x^2+y^2=4x2+y2=4.

sol:

Write D={(r,θ)∣1≤r≤2,0≤θ≤π}D = \{(r,\theta) | 1 \le r \le 2, 0 \le \theta \le \pi\}D={(r,θ)∣1≤r≤2,0≤θ≤π}.
∫∫D3x+4y2dA=∫12∫0π(3rcosθ+4r2sin2θ)rdθdr\int \int_D 3x+4y^2 dA = \int_1^2 \int_0^{\pi} (3rcos\theta + 4r^2sin^2\theta)r d\theta dr ∫∫D3x+4y2dA=∫12∫0π(3rcosθ+4r2sin2θ)rdθdr
E.x.

Compute the volume of the solid bounded by the plane z=0 and the paraboloid z=1−x2−y2z=1-x^2-y^2z=1−x2−y2.

sol:
∫∫D1−x2−y2dA=∫01∫02π(1−r2)rdθdr\int \int_D 1-x^2-y^2 dA = \int_0^1 \int_0^{2\pi} (1-r^2)r d\theta dr ∫∫D1−x2−y2dA=∫01∫02π(1−r2)rdθdr
E.x.

Find the region enclosed by r=cos2θr=cos2\thetar=cos2θ. For simplicity, restrict −π/4≤θ≤π/4-\pi/4 \le \theta \le \pi/4−π/4≤θ≤π/4.

sol:
area=∫π/4−π/4∫0cos2θrdrdθ=∫π/4−π/412(cos2θ)2dθ=∫π/4−π/4(1+cos4θ)4dθ=π/8area = \int_{\pi/4}^{-\pi/4} \int_{0}^{cos 2 \theta} r dr d\theta\\\\ = \int_{\pi/4}^{-\pi/4} \frac 1 2 (cos 2\theta)^2 d\theta \\\\ = \int_{\pi/4}^{-\pi/4} \frac{(1 + cos 4\theta)}{4} d\theta \\\\ = \pi/8 area=∫π/4−π/4∫0cos2θrdrdθ=∫π/4−π/421(cos2θ)2dθ=∫π/4−π/44(1+cos4θ)dθ=π/8

Applications of Double Integrals

Surface Area Element

S is a surface given by z=f(x,y). Derive a formula for computing the area of S. Assume that the domain of f is rectangular. The idea is to approximate S by tangent planes. Let ΔTij\Delta T_{ij}ΔTij to represent such planes. The work remaining is to compute ΔTij\Delta T_{ij}ΔTij.

Name the points of one such plane as P, Q, R, W. The vector PQ⃗=<Δx,0,fx(xi,yj)Δx>\vec{PQ} = <\Delta x, 0, f_x(x_i,y_j)\Delta x>PQ=<Δx,0,fx(xi,yj)Δx> , and the vector PR⃗=<0,Δy,fy(xi,yj)Δy>\vec{PR} = <0, \Delta y, f_y(x_i,y_j)\Delta y>PR=<0,Δy,fy(xi,yj)Δy> . The area of the parallelogram PQWR is given by
∣PQ⃗×PR⃗∣=ΔxΔy∣<−fx(xi,yj),−fy(xi,yj),1>∣|\vec{PQ} \times \vec{PR}| = \Delta x\Delta y|<-f_x(x_i,y_j),-f_y(x_i,y_j),1>| ∣PQ×PR∣=ΔxΔy∣<−fx(xi,yj),−fy(xi,yj),1>∣
Using Riemann sum, we obtain the area of S as follows
area(S)=∫∫Dfx2+fy2+1dxdyarea(S) = \int \int_D \sqrt{f_x^2 + f_y^2 + 1} dxdy area(S)=∫∫Dfx2+fy2+1dxdy

Density and Mass

The density at the point (x,y) is given by
ρ(x,y)=limΔmΔA\rho(x,y) = lim \frac{\Delta m}{\Delta A} ρ(x,y)=limΔAΔm
Assume we know that density ρ(x,y)\rho(x,y)ρ(x,y) and we would like to compute the mass of the object.
m=∫∫Dρ(x,y)dAm = \int \int_D \rho(x,y) dA m=∫∫Dρ(x,y)dA
The same idea can be applied to the charge density and total charge.

Center of Mass

The moment of the lamina about the x-axis and y-axis is
Mx=∫∫Dyρ(x,y)dAMy=∫∫Dxρ(x,y)dAM_x = \int\int_D y\rho(x,y) dA \\\\ M_y = \int\int_D x\rho(x,y) dA \\\\ Mx=∫∫Dyρ(x,y)dAMy=∫∫Dxρ(x,y)dA
The center of the mass is (xˉ,yˉ)(\bar x, \bar y)(xˉ,yˉ) so that mxˉ=Mym \bar x = M_ymxˉ=My and myˉ=Mxm \bar y = M_xmyˉ=Mx. Therefore,
xˉ=1m∫∫Dxρ(x,y)dAyˉ=1m∫∫Dyρ(x,y)dA\bar x = \frac 1 m \int\int_D x\rho(x,y) dA \\\\ \bar y = \frac 1 m \int\int_D y\rho(x,y) dA \\\\ xˉ=m1∫∫Dxρ(x,y)dAyˉ=m1∫∫Dyρ(x,y)dA
E.x.

The region is a half-circle with center at the origin and radius of a. The density of a lamina is x2+y2\sqrt{x^2+y^2}x2+y2. Find its center of mass.

sol:

Based on symmetry, we know MxM_xMx is 0.
My=∫∫Dyx2+y2dA=∫0π∫0ar3sinθdrdθM_y = \int\int_D y\sqrt{x^2+y^2} dA \\\\ = \int_0^{\pi}\int_0^a r^3 sin\theta \ drd\theta My=∫∫Dyx2+y2dA=∫0π∫0ar3sinθ drdθ

Probability

Let f(x,y) stand for the pmf.
P[(x,y)∈D]=∫∫Df(x,y)dAP[(x,y) \in D] = \int\int_D f(x,y) dA P[(x,y)∈D]=∫∫Df(x,y)dA

Triple Integrals

The Fubini’s Theorem can be extended to triple integrals.

E in R3R^3R3 is said to be of type I if it can be written as
E={(x,y,z)∣(x,y)∈D,D⊂R2,u1(x,y)≤z≤u2(x,y)}∫∫∫Ef(x,y,z)dV=∫∫D[∫u1(x,y)u2(x,y)f(x,y,z)dz]dAE = \{(x,y,z)|(x,y) \in D, D \sub R^2, u_1(x,y) \le z \le u_2(x,y) \} \\\\ \int\int\int_E f(x,y,z) dV = \int\int_D[\int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z) dz] dA E={(x,y,z)∣(x,y)∈D,D⊂R2,u1(x,y)≤z≤u2(x,y)}∫∫∫Ef(x,y,z)dV=∫∫D[∫u1(x,y)u2(x,y)f(x,y,z)dz]dA
If we also know that D is, say, type I, then the above triple is further equal to
∫ab∫g1(x)g2(x)∫u1(x,y)u2(x,y)f(x,y,z)dzdydx\int_a^b\int_{g_1(x)}^{g_2(x)}\int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z) dzdydx ∫ab∫g1(x)g2(x)∫u1(x,y)u2(x,y)f(x,y,z)dzdydx
E.x.

E is bounded by the coordinates and the plane x+y+z=1. Find ∫∫∫EzdV\int\int\int_E z dV∫∫∫EzdV.

sol:

Define the projection of E to the xy-plane to be D.
E={(x,y,z)∣(x,y)∈D,0≤z≤1−x−y}∫∫∫EzdV=∫01∫01−x∫01−x−yzdzdydxE = \{(x,y,z)|(x,y) \in D, 0 \le z \le 1-x-y \} \\\\ \int\int\int_E z dV = \int_0^1\int_{0}^{1-x}\int_{0}^{1-x-y} z dzdydx E={(x,y,z)∣(x,y)∈D,0≤z≤1−x−y}∫∫∫EzdV=∫01∫01−x∫01−x−yzdzdydx
E.x.

Use triple integrals to find the volume of the tetrahedron T bounded by the plane x+2y+z=2, x=2y, x=0, and z=0.

sol:

Draw the picture. Find the intersection. A=(0,0,2), B=(0,1,0), C=(1,1/2,0). We want to find volume of OABC.
VOABC=∫∫∫E1dV=∫01∫x/21−x/2∫02−x−2ydzdydxV_{OABC} = \int\int\int_E 1 dV \\\\ = \int_0^1\int_{x/2}^{1-x/2}\int_{0}^{2-x-2y} dzdydx \\\\ VOABC=∫∫∫E1dV=∫01∫x/21−x/2∫02−x−2ydzdydx

Triple Integrals in Cylindrical Coordinates

Say E is a type I and D has a clean form in the polar coordinate.
D={(r,θ)∣α≤θ≤β,h1(θ)≤r≤h2(θ)}D = \{(r,\theta)| \alpha \le \theta \le \beta, h_1(\theta) \le r \le h_2(\theta) \} D={(r,θ)∣α≤θ≤β,h1(θ)≤r≤h2(θ)}
It is just the polar coordinates adding the z-axis.
∫∫∫Ef(x,y,z)dV=∫∫∫Vf(r,θ,z)rdzdrdθ\int\int\int_E f(x,y,z) dV = \int\int\int_V f(r,\theta, z)r \ dz drd\theta ∫∫∫Ef(x,y,z)dV=∫∫∫Vf(r,θ,z)r dzdrdθ
E.x.

Describe the surface given by z=r in the cylindrical coordinate.

sol:
z=r=x2+y2z = r = \sqrt{x^2+y^2} z=r=x2+y2
E.x.

A solid E lies within the cylinder x2+y2=1x^2+y^2=1x2+y2=1, below the plane z=4 and above the paraboloid z=1−(x2+y2)z=1-(x^2+y^2)z=1−(x2+y2). The density function is x2+y2\sqrt{x^2+y^2}x2+y2. Find its mass.

sol:
∫∫∫Eρ(x,y,z)dV=∫02π∫01∫1−r24r2dzdrdθ\int\int\int_E \rho(x,y,z) dV \\\\ = \int_0^{2\pi}\int_0^1\int_{1-r^2}^{4} r^2 \ dzdrd\theta \\\\ ∫∫∫Eρ(x,y,z)dV=∫02π∫01∫1−r24r2 dzdrdθ

Triple Integrals in Spherical Coordinates

We use (ρ,θ,ϕ)(\rho, \theta, \phi)(ρ,θ,ϕ) to represent a point P in the spherical coordinates. The first component is the distance between O and P. The third component is the angle between OP⃗\vec{OP}OP and the positive direction of z-axis (o≤ϕ≤πo \le \phi \le \pio≤ϕ≤π).
x=ρsinϕcosθy=ρsinϕsinθz=ρcosϕx = \rho sin \phi cos \theta \\\\ y = \rho sin \phi sin \theta \\\\ z = \rho cos \phi \\\\ x=ρsinϕcosθy=ρsinϕsinθz=ρcosϕ
The formula is as below
∫∫∫Ef(x,y,z)dV=∫cd∫αβ∫abf(ρ,θ,ϕ)ρ2sinϕdρdθdϕ\int\int\int_E f(x,y,z) dV = \int_{c}^{d} \int_{\alpha}^{\beta} \int_{a}^{b} f(\rho, \theta, \phi) \rho^2 sin \phi \ d\rho d \theta d \phi ∫∫∫Ef(x,y,z)dV=∫cd∫αβ∫abf(ρ,θ,ϕ)ρ2sinϕ dρdθdϕ
E.x.

Compute the volume between the cone z=x2+y2z=\sqrt{x^2+y^2}z=x2+y2 and the sphere x2+y2+z2=zx^2+y^2+z^2=zx2+y2+z2=z.

sol:

The latter part is a sphere of radius 0.5 centered at (0,0,0.5). In the spherical coordinates, the two planes can be expressed as
cosϕ=sinϕρ=cosϕcos \phi = sin \phi\\\\ \rho = cos \phi \\\\ cosϕ=sinϕρ=cosϕ
The solid can be expressed as
E={(ρ,θ,ϕ)∣0≤ϕ≤π/4,0≤ρ≤cosϕ,0≤θ≤2π}E = \{(\rho, \theta, \phi)| 0 \le \phi \le \pi/4, 0 \le \rho \le cos \phi, 0 \le \theta \le 2\pi\} E={(ρ,θ,ϕ)∣0≤ϕ≤π/4,0≤ρ≤cosϕ,0≤θ≤2π}

Change of Variables in Multiple Integrals

One-to-One Maps

Change of variable for integrals.

We have f:[a,b]−>Rf:[a,b]->Rf:[a,b]−>R and want to substitute x=g(u)x=g(u)x=g(u). Let g(c)=a,g(d)=bg(c)=a, g(d)=bg(c)=a,g(d)=b. We need one assumption that g is monotonic.
∫abf(x)dx=∫cdf(g(u))g′(u)du\int_a^b f(x)dx = \int_c^d f(g(u))g'(u) du ∫abf(x)dx=∫cdf(g(u))g′(u)du
Change of variable for double integrals. Special case: polar coordinate.

In general, transformation T(u,v)=(x,y)T(u,v) = (x,y)T(u,v)=(x,y). (u,v) and (x,y) are connected by
x=g(u,v)y=h(u,v)x = g(u,v)\\\\ y = h(u,v) x=g(u,v)y=h(u,v)
If it happens that for (u1,v1)≠(u2,v2)(u_1, v_1) \ne (u_2, v_2)(u1,v1)=(u2,v2) and T(u1,v1)=T(u2,v2)T(u_1, v_1) = T(u_2, v_2)T(u1,v1)=T(u2,v2) then we say that the map T is not one-to-one.

Given (x2,y2)(x_2, y_2)(x2,y2), we can find one (u2,v2)(u_2, v_2)(u2,v2) s.t. T(u2,v2)=(x2,y2)T(u_2, v_2)=(x_2, y_2)T(u2,v2)=(x2,y2). This is called T−1T^{-1}T−1.

To see whether a map is one-to-one:

T sends different points to different points
for every (x,y)∈R2(x,y)\in R^2(x,y)∈R2, we are able to find (u,v)(u, v)(u,v) s.t. T(u,v)=(x,y)T(u, v)=(x, y)T(u,v)=(x,y).

Change Variable Formula

We approximate the region after transformation by a parallelogram.

The area of such parallelogram is
AB⃗=T(u+Δu,v)−T(u,v)=<g(u+Δu,v)−g(u,v),h(u+Δu,v)−h(u,v)>=<guΔu,huΔu>AC⃗=T(u,v+Δv)−T(u,v)=<g(u,v+Δv)−g(u,v),h(u,v+Δv)−h(u,v)>=<gvΔv,hvΔv>S=∣AB⃗×AC⃗∣=∣guhv−gvhu∣\vec{AB} = T(u+\Delta u, v) - T(u, v) \\\\ = <g(u+\Delta u, v) - g(u,v), h(u+\Delta u, v) - h(u,v)> \\\\ = <g_u \Delta u, h_u \Delta u> \\\\ \vec{AC} = T(u, v+\Delta v) - T(u, v) \\\\ = <g(u, v+\Delta v) - g(u,v), h(u, v+\Delta v) - h(u,v)> \\\\ = <g_v \Delta v, h_v \Delta v> \\\\ S = |\vec{AB} \times \vec{AC}| = |g_uh_v- g_vh_u| AB=T(u+Δu,v)−T(u,v)=<g(u+Δu,v)−g(u,v),h(u+Δu,v)−h(u,v)>=<guΔu,huΔu>AC=T(u,v+Δv)−T(u,v)=<g(u,v+Δv)−g(u,v),h(u,v+Δv)−h(u,v)>=<gvΔv,hvΔv>S=∣AB×AC∣=∣guhv−gvhu∣
Definition of Jacobian of the transformation T

Given by x=g(u,v)x = g(u,v)x=g(u,v) and y=h(u,v)y = h(u,v)y=h(u,v) is
∂(x,y)∂(u,v)=guhv−gvhu\frac{\partial (x,y)}{\partial (u,v)} = g_uh_v- g_vh_u ∂(u,v)∂(x,y)=guhv−gvhu
In the general case, the change of variable in integrals
∫∫∫EdV=∫∫∫V∣∂(x,y,z)∂(u,v,w)∣dudvdw\int\int\int_E dV = \int\int\int_V |\frac{\partial(x,y,z)}{\partial(u,v,w)}| \ dudvdw ∫∫∫EdV=∫∫∫V∣∂(u,v,w)∂(x,y,z)∣ dudvdw
E.x.

We have a region R bounded by (1,0), (2,0), (0,-2), and (0,-1). Evaluate ∫∫Rex+yx−ydA\int\int_R e^{\frac{x+y}{x-y}} dA∫∫Rex−yx+ydA.

sol:

We first let u=x+y,v=x−yu=x+y, \ v=x-yu=x+y, v=x−y. Verify that this is an one-to-one map.

Compute the Jacobian.
∂(x,y)∂(u,v)=xuyv−xvyu=−12\frac{\partial (x,y)}{\partial (u,v)} = x_uy_v- x_vy_u = -\frac 1 2 ∂(u,v)∂(x,y)=xuyv−xvyu=−21
The next step is to figure out how R is transformed under the map T.

(x,y) = (1,0) => (u,v) = (1,1)

(x,y) = (2,0) => (u,v) = (2,2)

(x,y) = (0,-1) => (u,v) = (-1,1)

(x,y) = (0,-2) => (u,v) = (-2,2)
∫∫Rex+yx−ydA=∫∫Seuv∣−12∣dudv=∫12∫−vv12euvdudv=∫12[12veuv]−vvdv\int\int_R e^{\frac{x+y}{x-y}} dA = \int\int_S e^{\frac{u}{v}} |-\frac 1 2| dudv \\\\ = \int_{1}^{2}\int_{-v}^{v} \frac 1 2 e^{\frac{u}{v}} dudv \\\\ = \int_{1}^{2} [\frac 1 2 ve^{\frac u v}]_{-v}^{v} dv \\\\ ∫∫Rex−yx+ydA=∫∫Sevu∣−21∣dudv=∫12∫−vv21evududv=∫12[21vevu]−vvdv

Reference

Multivariable_Calculus_8th_Edition (15.1-15.9), James Stewart

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Calc3: Multiple Integrals

Double Integrals Over Rectangles

Geometry

Iterated Integrals

Double Integrals Over General Regions

Double Integrals In Polar Coordinates

Applications of Double Integrals

Surface Area Element

Density and Mass

Center of Mass

Probability

Triple Integrals

Triple Integrals in Cylindrical Coordinates

Triple Integrals in Spherical Coordinates

Change of Variables in Multiple Integrals

One-to-One Maps

Change Variable Formula

Reference

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