Multiple integrals and applications notes — Unit 4

A single integral ∫ f(x) dx finds the area under a 2D curve. But what if you have a 3D surface z = f(x,y) and you want to find the volume underneath it? You need a Double Integral: ∬ f(x,y) dx dy. And what if you want to find the mass of a solid block where the density varies at every point? You need a Triple Integral: ∭ f(x,y,z) dx dy dz.

⇒The Geometry of Limits

In a single integral, limits are just two numbers on the x-axis: x=a to x=b. In a double integral, the limits define a 2D region (an area) on the xy-plane.

To evaluate ∬ f(x,y) dx dy, we work from the INSIDE OUT.

Evaluate I = ∫(x=0 to 1) ∫(y=0 to x) (x² + y²) dy dx

1. 1.Step 1: The inner integral is w.r.t 'y' (from y=0 to y=x). Treat x as a constant.
2. 2.Inner Integral = ∫ (x² + y²) dy
3. 3. = [ x²y + y³/3 ] evaluated from y=0 to y=x
4. 4. = ( x²(x) + x³/3 ) - ( 0 + 0 )
5. 5. = x³ + x³/3
6. 6. = (4/3)x³

Step 2: Plug this back into the outer integral w.r.t 'x'.
I = ∫(x=0 to 1) (4/3)x³ dx
  = (4/3) [ x⁴ / 4 ] from 0 to 1
  = (4/3) * (1/4 - 0)
  = 1/3

Often, you are not given the explicit limits. You are told: 'Integrate f(x,y) over the region bounded by y=x and y=x²'. You must sketch the region and extract the limits yourself.

Always draw a tiny representative strip (rectangle) on your sketch. This single habit prevents 90% of exam errors.

This is the most guaranteed 10-mark question in Unit 4. You will be given an integral that is mathematically impossible to evaluate as written (e.g., integrating e^(y²) dy). The ONLY way to solve it is to change the order from dy dx to dx dy.

Change the order and evaluate: I = ∫(y=0 to ∞) ∫(x=y to ∞) [ e^(-x) / x ] dx dy

1. 1.Attempting to evaluate directly: ∫ e^(-x)/x dx has no closed-form elementary anti-derivative. It's impossible. We must change order.
2. 2.Step 1: Extract old limits.
3. 3.Inner (dx): x varies from x = y to x = ∞
4. 4.Outer (dy): y varies from y = 0 to y = ∞

Step 2: Sketch the region.
The line x = y is a 45-degree line passing through origin.
y = 0 is the x-axis.
The region goes from the line x=y out to x=∞ (the right side).
And from y=0 up to y=∞.
This means the region is the entire infinite wedge between the x-axis and the line y=x in the first quadrant.

Step 3: Rotate the strip.

Old order (dx dy) used HORIZONTAL strips (left to right).
The left end touched x = y. The right end went to x = ∞.

New order (dy dx) will use VERTICAL strips (bottom to top).
Look at the wedge region. For any vertical strip:
The BOTTOM end touches the x-axis (y = 0).
The TOP end touches the line (y = x).

1. 1.Step 4: Extract new limits.
2. 2.Inner (dy): y varies from y = 0 to y = x.
3. 3.Outer (dx): To cover the whole infinite wedge, x must scan from x = 0 all the way to x = ∞.

New I = ∫(x=0 to ∞) ∫(y=0 to x) [ e^(-x) / x ] dy dx

Now let's evaluate the newly ordered integral.

New I = ∫(x=0 to ∞) ∫(y=0 to x) [ e^(-x) / x ] dy dx

Step 1: Evaluate inner integral w.r.t 'y'.
The term [e^(-x)/x] acts entirely as a constant!
∫ [ e^(-x) / x ] dy = [ e^(-x) / x ] * [ y ] evaluated from 0 to x
                    = [ e^(-x) / x ] * (x - 0)
                    = e^(-x)

1. 1.This is amazing. The annoying 'x' in the denominator completely canceled out with the 'x' limit from the inner integration.
2. 2.Step 2: Evaluate outer integral w.r.t 'x'.
3. 3.I = ∫(x=0 to ∞) e^(-x) dx
4. 4. = [ -e^(-x) ] from 0 to ∞
5. 5. = ( -e^(-∞) ) - ( -e^(0) )
6. 6. = ( 0 ) - ( -1 )
7. 7. = 1

Sometimes Cartesian (x,y) coordinates are terrible for the region. If the region is a circle x² + y² = a², vertical strips will have limits involving square roots: y = ±√(a² - x²). Integrating this is brutal. We switch to Polar Coordinates (r, θ).

J(x,y / r,θ) = | ∂x/∂r  ∂x/∂θ | = | cosθ  -rsinθ |
               | ∂y/∂r  ∂y/∂θ |   | sinθ   rcosθ |

J = r cos²θ - (-r sin²θ) = r(cos²θ + sin²θ) = r

Therefore: dx dy = r dr dθ

Evaluate ∬ e^(-(x²+y²)) dx dy over the positive quadrant of a circle of radius R.

1. 1.1. Transform variables: x² + y² becomes r².
2. 2.2. Transform differential: dx dy becomes r dr dθ.
3. 3.3. Transform region limits:
4. 4. The region is a quarter circle (first quadrant).
5. 5. r (distance from origin) goes from 0 to R.
6. 6. θ (angle from x-axis) goes from 0 to π/2.

I = ∫(θ=0 to π/2) ∫(r=0 to R) e^(-r²) * r dr dθ

Inner Integral (dr):
Let u = r², then du = 2r dr  => r dr = du/2
∫ e^(-u) du/2 = (-1/2) e^(-r²) evaluated from 0 to R
              = (-1/2) (e^(-R²) - 1)
              = (1 - e^(-R²)) / 2

Outer Integral (dθ):
I = ∫(0 to π/2) [ (1 - e^(-R²)) / 2 ] dθ
  = [ (1 - e^(-R²)) / 2 ] * [ θ ] from 0 to π/2
  = (π/4) * (1 - e^(-R²))

Triple Integrals ∭ f(x,y,z) dx dy dz follow the exact same logic as double integrals. Work inside out. The innermost integral has limits depending on 2 variables. The middle integral has limits depending on 1 variable. The outermost integral has constant limits.

Always look at the region of integration. If it has x² + y² + z² in it, switch to spherical immediately. If it has x² + y², switch to cylindrical.

Just like dx dy = r dr dθ required a Jacobian, switching a triple integral requires a 3x3 Jacobian.

x = r cosθ,  y = r sinθ,  z = z
Jacobian J = r
dx dy dz = r dr dθ dz

x = r sinθ cosφ
y = r sinθ sinφ
z = r cosθ
Jacobian J = r² sinθ  (Derived in Unit 3 notes!)
dx dy dz = r² sinθ dr dθ dφ

To find the pure area of a 2D region, simply evaluate the double integral with f(x,y) = 1.

Area = ∬ 1 dx dy

1. 1.Example: Find area between parabola y = x² and line y = x.
2. 2.1. Find intersections: x² = x => x(x-1) = 0 => x=0, x=1.
3. 3.2. Set limits: x goes 0 to 1. For a given x, y goes from x² (bottom curve) to x (top curve).
4. 4.3. Area = ∫(x=0 to 1) ∫(y=x² to x) 1 dy dx
5. 5.4. Inner: [y] from x² to x = (x - x²)
6. 6.5. Outer: ∫(x-x²) dx = [x²/2 - x³/3] from 0 to 1 = 1/2 - 1/3 = 1/6.

To find the pure volume of a 3D region, simply evaluate the triple integral with f(x,y,z) = 1.

Volume = ∭ 1 dx dy dz

Alternatively, if you know the ceiling function z = f(x,y) and the floor is z=0, you can use a double integral.

Volume = ∬ f(x,y) dx dy

Prove the volume of a sphere of radius R is (4/3)πR³.

Equation: x² + y² + z² = R²
Using Cartesian is horrible. Use Spherical coordinates.
Volume = ∭ 1 dx dy dz = ∭ (r² sinθ) dr dθ dφ

1. 1.Set limits for a full sphere:
2. 2.r (radius) goes from 0 to R
3. 3.θ (latitude) goes from 0 to π (North pole to South pole)
4. 4.φ (longitude) goes from 0 to 2π (Full circle around)

V = ∫(φ=0 to 2π) ∫(θ=0 to π) ∫(r=0 to R) (r² sinθ) dr dθ dφ

Because the limits are all constants and the terms are multiplied, we can split this into 3 independent integrals!

V = [ ∫(0 to 2π) dφ ] * [ ∫(0 to π) sinθ dθ ] * [ ∫(0 to R) r² dr ]
V = [ 2π ] * [ -cos(π) - (-cos(0)) ] * [ R³/3 ]
V = [ 2π ] * [ 1 - (-1) ] * [ R³/3 ]
V = [ 2π ] * [ 2 ] * [ R³/3 ]
V = (4/3)πR³

If a solid has a non-uniform density, say ρ(x,y,z) = x²yz, you cannot just say Mass = Volume * Density. You must integrate the density over the volume.

These are long, tedious calculations. In exams, look for symmetry! If the object is perfectly symmetric across the yz-plane, and the density is symmetric, then x̄ = 0 instantly by inspection.

Dirichlet's Theorem is a powerful shortcut for a very specific type of triple integral that appears constantly in exams.

Find the mass of a solid in the first octant bounded by x+y+z=1, if density ρ = xyz.

1. 1.Step 1: Set up the integral.
2. 2.Mass = ∭ (xyz) dx dy dz
3. 3.Step 2: Match to Dirichlet form: x^(l-1) y^(m-1) z^(n-1)
4. 4.Here, x = x¹. So l-1 = 1 => l=2.
5. 5.Similarly, m=2, n=2.

Step 3: Apply the formula directly.
Mass = [ Γ(2) * Γ(2) * Γ(2) ] / Γ(2 + 2 + 2 + 1)
     = [ Γ(2)³ ] / Γ(7)
     = [ (1!)³ ] / 6!
     = 1 / 720

We solved a mass integration problem over a tetrahedral region in 3 lines of algebra without doing a single integration step. This is the power of Dirichlet.

What if the region is bounded by (x/a)^p + (y/b)^q + (z/c)^r ≤ 1? Dirichlet's standard theorem fails because of the powers and constants.

1. 1.Liouville allows us to use substitution to 'force' it into Dirichlet form.
2. 2.Let u = (x/a)^p => x = a * u^(1/p)
3. 3.Then dx = (a/p) * u^(1/p - 1) du
4. 4.Do the same for y (into v) and z (into w).
5. 5.The region is now strictly u + v + w ≤ 1.

You can now plug the new dx, dy, dz into the integral and apply Dirichlet.
This technique is heavily used to find the volume of an Ellipsoid (x²/a² + y²/b² + z²/c² = 1).

⇒Unit 4 Mastery

• Can you draw horizontal and vertical strips to extract limits from a 2D sketch?
• Do you remember the 4-step algorithm to Change the Order of Integration?
• Can you correctly evaluate a double integral from the INSIDE OUT?
• Do you remember the Jacobians for Polar (r), Cylindrical (r), and Spherical (r²sinθ)?
• Can you set up a triple integral to find Volume and Mass?
• Can you execute Dirichlet's formula for tetrahedral boundary conditions?