At the origin, the limit of f(x, y) = xy/(x² + y²) as (x, y) → (0, 0) is:

nonexistent, because different lines give different values

1/2, because f = 1/2 along y = x

nonexistent, because different lines give different values

For f(x, y) = x y²/(x² + y⁴) at the origin, which statement is correct?

the limit does not exist, since the value is 0 along lines but 1/2 along x = y²

the limit is 0 because it is 0 along every straight line through the origin

the limit is 1/2 because f = 1/2 along the parabola x = y²

the limit does not exist, since the value is 0 along lines but 1/2 along x = y²

the limit is infinite along x = y², so f is unbounded

Which statements about functions of two variables are correct?

If |f(x, y)| ≤ |x| + |y| and f(0, 0) = 0 then f is continuous at the origin

If two paths give different limits then the limit does not exist

If both partial derivatives exist at a point then f is continuous there

If the limit is 0 along y = x and along y = −x then the limit exists and is 0

If |f(x, y)| ≤ |x| + |y| and f(0, 0) = 0 then f is continuous at the origin

If two paths give different limits then the limit does not exist

Which statements about the method of Lagrange multipliers are correct?

At a constrained extremum the gradient of f is parallel to the gradient of the constraint function

The solutions of the equations are candidates, so f must be compared at all of them

At a constrained extremum the gradient of f is parallel to the gradient of the constraint function

The solutions of the equations are candidates, so f must be compared at all of them

The multiplier λ must be positive at a maximum

On the line x + y = 10 over all real x, f = xy has a minimum value

Improper Integrals, Volumes of Revolution, Two-Variable Limits, Lagrange Multipliers and Double-Integral Applications

This chapter belongs to Section XE0 (Engineering Mathematics) of the Engineering Sciences (XE) paper, the part common to all candidates. It supplies the rest of XE0.2 that the shared chapters leave out: “Evaluation of definite and improper integrals; Applications of definite integrals to evaluate areas and volumes (rotation of a curve about an axis)” and, for functions of two variables, “Limit, continuity … Method of Lagrange multipliers; Double integrals and their applications.” Partial derivatives, maxima and minima, directional derivatives and the evaluation of double integrals are taught in the shared multivariable chapter and are not repeated. The numericals are an improper integral by the gamma function, volumes of solids of revolution, a constrained maximum, a constrained distance and the mass of a lamina.

1. Improper integrals

An integral is improper if an end of the interval is infinite (type I) or the integrand is unbounded at a point of the interval (type II). Each is defined as a limit: ∫ₐ^∞ f dx = lim_b→∞ ∫ₐᵇ f dx, and for f unbounded at a, ∫ₐᵇ f dx = lim_ε→0⁺ ∫_a+εᵇ f dx. The integral converges if the limit is finite. For a singularity inside the interval split at the singular point and require both pieces to converge.

The two benchmark integrals
Integral	Converges when	Value
∫₁^∞ dx/xᵖ	p > 1	1/(p − 1)
∫₀¹ dx/xᵖ	p < 1	1/(1 − p)

Compare with these: if 0 ≤ f ≤ g and ∫g converges then ∫f converges, and if f/g → c ∈ (0, ∞) the two behave alike. Examples. ∫₀^∞ e^(−x) dx = 1. ∫_−∞^∞ dx/(1 + x²) = π. ∫₀¹ ln x dx = [x ln x − x]₀¹ = −1. ∫₀^∞ xⁿ e^(−ax) dx = n!/aⁿ⁺¹ (the gamma function, Γ(n + 1) = n!), so ∫₀^∞ x³e^(−2x) dx = 3!/2⁴ = 3/8. ∫₀^∞ e^(−x²) dx = √π/2 is a standard value.

⚠️ Never cancel across a singularity

∫₋₁¹ dx/x² is not [−1/x]₋₁¹ = −2: the integrand is positive, and each half ∫₀¹ dx/x² diverges to +∞. The 'principal value' of ∫₋₁¹ dx/x is 0 by symmetry, but the integral itself diverges.

2. Areas and volumes of revolution

Rotate the region under y = f(x), a ≤ x ≤ b, about the x-axis. Slicing perpendicular to the axis gives discs of radius f(x): V = π∫ₐᵇ y² dx. If the region lies between y = f₁ ≥ f₂, the slices are washers: V = π∫(f₁² − f₂²) dx. For rotation about the y-axis of the region under y = f(x) the shell method gives V = 2π∫ₐᵇ x·y dx (0 ≤ a < b); equivalently use discs in y.

Worked. (i) y = √x, 0 ≤ x ≤ 4, about the x-axis: V = π∫₀⁴ x dx = π·16/2 = 8π. (ii) y = x², 0 ≤ x ≤ 1, about the y-axis: V = 2π∫₀¹ x·x² dx = 2π/4 = π/2; check with washers π∫₀¹ (1 − y) dy = π/2. (iii) The region between y = x and y = x² about the x-axis: V = π∫₀¹ (x² − x⁴) dx = π(1/3 − 1/5) = 2π/15. (iv) Circle x² + y² = a² about the x-axis: V = π∫₋ₐᵃ (a² − x²) dx = 4πa³/3, the sphere.

Areas. Between curves, the area is ∫(upper − lower) dx, splitting wherever the curves cross (shared chapter). In polar form, the area swept by r = r(θ) from α to β is A = ½∫ₐᵝ r² dθ. The cardioid r = a(1 + cos θ) has A = ½a²∫₀^2π (1 + cos θ)² dθ = ½a²(2π + π) = 3πa²/2, since ∫cos² θ dθ over a full turn is π and the cross term integrates to 0.

⚠️ Axis and variable must match

Rotation about the x-axis integrates y² dx; rotation of the same curve about the y-axis is a different solid (use shells x·y dx). For the shell formula the strip distance from the axis is x, so the region must not cross the axis.

3. Limits and continuity for two variables

lim_{(x,y)→(a,b)} f = L means f is close to L whenever (x, y) is close to (a, b) from every direction and along every path. So if two paths give different values, the limit does not exist; agreement on many paths proves nothing by itself. f is continuous at (a, b) if the limit exists and equals f(a, b).

Examples at the origin. xy/(x² + y²): along y = mx the value is m/(1 + m²), which depends on m, so no limit. xy²/(x² + y⁴): along every line y = mx it tends to 0, but along x = y² it is y⁴/(2y⁴) = 1/2, so no limit. x²y/(x² + y²): |x²y/(x² + y²)| ≤ |y| → 0, so the limit is 0 (squeeze). In polar form x = r cos θ, y = r sin θ the limit exists if the expression is bounded by something like r·g(θ) with g bounded; (x³ + y³)/(x² + y²) = r(cos³θ + sin³θ) → 0.

⚠️ Partial derivatives can exist without continuity

Define f = xy/(x² + y²) with f(0, 0) = 0. Both f_x(0, 0) and f_y(0, 0) equal 0 because f vanishes on both axes, yet f is not continuous at the origin. In one variable differentiable means continuous; in two variables the existence of partial derivatives is weaker than continuity. Continuous partials do imply differentiability.

4. Method of Lagrange multipliers

To find extrema of f(x, y) subject to a constraint g(x, y) = 0, solve ∇f = λ∇g together with g = 0. At a constrained extremum the level curve of f is tangent to the constraint curve, so the two gradients are parallel; λ is the multiplier. With three variables the same equations hold with ∇ in three components. The method only gives candidates: evaluate f at each and compare; on a closed bounded constraint set (a circle, an ellipse) a maximum and a minimum are guaranteed to exist.

Worked. (i) Maximise f = x + 2y on x² + y² = 5: 1 = 2λx and 2 = 2λy give y = 2x; then 5x² = 5, x = ±1. At (1, 2): f = 5 (maximum); at (−1, −2): f = −5 (minimum). (ii) Minimise x² + y² on x + y = 2: 2x = λ = 2y so x = y = 1 and the minimum is 2 (the squared distance from the origin to the line, √2 away). (iii) Maximise f = xy² on x + y = 9 with x, y ≥ 0: y² = λ and 2xy = λ give y = 2x (for y ≠ 0), so 3x = 9, x = 3, y = 6 and f = 3 × 36 = 108.

⚠️ Check the endpoints and the other candidates

On x + y = 9 the product f = xy² has no maximum if x, y may be any real number (it grows without bound), so the domain statement x, y ≥ 0 is part of the problem. Dividing both equations by y silently drops the candidate y = 0; there f = 0, which is a minimum, not the maximum.

5. Applications of double integrals

What a double integral computes
Quantity	Formula over the region R
Area	A = ∬ dA
Volume under z = f(x, y)	V = ∬ f dA
Mass, surface density ρ	M = ∬ ρ dA
Centroid (x̄, ȳ)	x̄ = ∬ xρ dA / M, ȳ = ∬ yρ dA / M
Moment of inertia about the z-axis	I_z = ∬ ρ(x² + y²) dA

For a change of variables (x, y) → (u, v), dA = |∂(x, y)/∂(u, v)| du dv, where the Jacobian is the determinant of the matrix of partial derivatives; in polar form it is r, giving dA = r dr dθ. Worked. (i) The paraboloid z = 4 − x² − y² above the plane z = 0: in polar form V = ∫₀^2π∫₀² (4 − r²) r dr dθ = 2π(8 − 4) = 8π. (ii) A uniform disc of mass M and radius a: ρ = M/(πa²), so I_z = ρ·2π∫₀ᵃ r³ dr = ρ·2π·a⁴/4 = Ma²/2. (iii) A triangle with vertices (0, 0), (a, 0), (0, a), uniform: x̄ = ȳ = a/3.

Mass of a variable-density lamina. For ρ = k(x + y) on the triangle x, y ≥ 0, x + y ≤ 1: M = k∫₀¹∫₀^1−x (x + y) dy dx = k∫₀¹ [x(1 − x) + (1 − x)²/2] dx = k(1/6 + 1/6) = k/3. With k = 12 kg/m² per metre the mass is 4 kg.

🧠 Sketch, then choose the order

Describe R first (limits of the inner variable in terms of the outer), then integrate. A circle, annulus or sector is always easier in polar form; do not forget the factor r.

Key takeaways

Improper integrals are limits: ∫₁^∞ x⁻ᵖ converges iff p > 1, ∫₀¹ x⁻ᵖ iff p < 1; ∫₀^∞ xⁿe^(−ax) = n!/aⁿ⁺¹; never cancel across a singularity.
Volume of revolution about the x-axis is π∫y² dx (washers for a gap); about the y-axis use shells 2π∫xy dx; polar area is ½∫r² dθ.
A two-variable limit exists only if every path gives the same value; two different path values disprove it, and polar form or squeezing proves it.
Existence of partial derivatives does not imply continuity; continuous partials imply differentiability.
Lagrange multipliers: solve ∇f = λ∇g and g = 0, then compare f at every candidate; mind domain restrictions and lost solutions.
Double integrals give area, volume, mass, centroid and I = ∬ρr² dA; use dA = r dr dθ in polar form and the Jacobian in general.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The integral ∫₁^∞ dx/xᵖ converges for:
1. p ≥ 1
2. p < 1
3. p > 1
4. every real p
Show answer
Answer: C — p > 1
The antiderivative is x^(1−p)/(1 − p), which tends to 0 as x → ∞ only if 1 − p < 0, i.e. p > 1, and then the value is 1/(p − 1). At p = 1 the integral is ln x → ∞, so p ≥ 1 is wrong.
The value of the integral ∫₋₁¹ dx/x² is:
1. it diverges to +∞
2. −2
3. 0
4. 2
Show answer
Answer: A — it diverges to +∞
The integrand is positive and unbounded at x = 0, and ∫₀¹ dx/x² has p = 2 ≥ 1 so it diverges. Applying [−1/x] across 0 to get −2 is invalid, since the antiderivative is discontinuous there and a positive function cannot have a negative integral.
Which of the following improper integrals is convergent?
1. ∫₁^∞ dx/√x
2. ∫₀¹ dx/x
3. ∫₂^∞ dx/(x ln x)
4. ∫₀¹ dx/√x
Show answer
Answer: D — ∫₀¹ dx/√x
∫₀¹ x^(−1/2) dx = [2√x]₀¹ = 2, convergent since p = 1/2 < 1. ∫₁^∞ x^(−1/2) dx diverges (p < 1 at infinity), ∫₀¹ dx/x diverges (p = 1), and ∫dx/(x ln x) = ln ln x → ∞.
The region under y = √x for 0 ≤ x ≤ 4 is rotated about the x-axis. The volume of the solid is:
1. 16π
2. 8π
3. 4π
4. 32π/3
Show answer
Answer: B — 8π
V = π∫₀⁴ y² dx = π∫₀⁴ x dx = π·[x²/2]₀⁴ = 8π. 16π results from forgetting the factor 1/2 in the integral of x; integrating y instead of y² would give π·16/3, which is not among the choices.
The region bounded by y = x², the x-axis and x = 1 is rotated about the y-axis. The volume generated is:
1. π/2
2. π/5
3. 2π/3
4. π/3
Show answer
Answer: A — π/2
Shells of radius x and height x²: V = 2π∫₀¹ x·x² dx = 2π/4 = π/2. Washers agree: π∫₀¹ (1 − y) dy = π/2. π/5 is the volume about the x-axis, π∫₀¹ x⁴ dx, a different solid.
At the origin, the limit of f(x, y) = xy/(x² + y²) as (x, y) → (0, 0) is:
1. 0, because f vanishes on both axes
2. 1/2, because f = 1/2 along y = x
3. 1, by L'Hospital's rule
4. nonexistent, because different lines give different values
Show answer
Answer: D — nonexistent, because different lines give different values
Along y = mx, f = m/(1 + m²): 0 for m = 0, 1/2 for m = 1, −1/2 for m = −1. Different paths give different values, so the limit does not exist. Vanishing on the axes, or agreeing 1/2 on one line, shows only that those paths do not decide it.
For f(x, y) = x y²/(x² + y⁴) at the origin, which statement is correct?
1. the limit is 0 because it is 0 along every straight line through the origin
2. the limit is 1/2 because f = 1/2 along the parabola x = y²
3. the limit does not exist, since the value is 0 along lines but 1/2 along x = y²
4. the limit is infinite along x = y², so f is unbounded
Show answer
Answer: C — the limit does not exist, since the value is 0 along lines but 1/2 along x = y²
Along y = mx: m²x³/(x² + m⁴x⁴) → 0. Along x = y²: y⁴/(y⁴ + y⁴) = 1/2. Two values, so no limit, even though every straight line gives 0. A single non-linear path is enough to disprove existence.
The minimum value of x² + y² subject to the constraint x + y = 2 is:
1. 4
2. 2
3. 1
4. √2
Show answer
Answer: B — 2
∇f = λ∇g gives 2x = λ and 2y = λ, so x = y, and the constraint then gives x = y = 1 and f = 2. The distance from the origin to the line is √2, but the quantity asked is its square, so √2 is wrong.
A uniform thin disc of mass M and radius a lies in the xy-plane. Using I = ∬ ρ (x² + y²) dA, its moment of inertia about the z-axis through the centre is:
1. Ma²
2. Ma²/4
3. 2Ma²/5
4. Ma²/2
Show answer
Answer: D — Ma²/2
In polar form I = ρ∫₀^2π∫₀ᵃ r²·r dr dθ = ρ·2π·a⁴/4 = (M/πa²)(πa⁴/2) = Ma²/2. Ma²/4 is the value about a diameter (perpendicular axes theorem), 2Ma²/5 is for a solid sphere and Ma² for a thin ring.
Which of the following improper integrals converge?
1. ∫₀^∞ e^(−3x) dx
2. ∫₁^∞ dx/√x
3. ∫₀¹ dx/√x
4. ∫₀¹ dx/x
Show answer
Answer: A — ∫₀^∞ e^(−3x) dx; C — ∫₀¹ dx/√x
∫₀^∞ e^(−3x) dx = 1/3 and ∫₀¹ x^(−1/2) dx = 2 converge. ∫₁^∞ x^(−1/2) dx = [2√x] → ∞ diverges (p = 1/2 ≤ 1), and ∫₀¹ dx/x = [ln x] → −∞ at 0 diverges (p = 1).
Which statements about functions of two variables are correct?
1. If both partial derivatives exist at a point then f is continuous there
2. If the limit is 0 along y = x and along y = −x then the limit exists and is 0
3. If |f(x, y)| ≤ |x| + |y| and f(0, 0) = 0 then f is continuous at the origin
4. If two paths give different limits then the limit does not exist
Show answer
Answer: C — If |f(x, y)| ≤ |x| + |y| and f(0, 0) = 0 then f is continuous at the origin; D — If two paths give different limits then the limit does not exist
The squeeze 0 ≤ |f| ≤ |x| + |y| → 0 gives continuity, and two differing path limits disprove existence. Existence of partials does not give continuity (xy/(x² + y²)), and agreement on two lines proves nothing (xy²/(x² + y⁴) is 0 on every line and 1/2 on x = y²).
Which statements about the method of Lagrange multipliers are correct?
1. At a constrained extremum the gradient of f is parallel to the gradient of the constraint function
2. The solutions of the equations are candidates, so f must be compared at all of them
3. The multiplier λ must be positive at a maximum
4. On the line x + y = 10 over all real x, f = xy has a minimum value
Show answer
Answer: A — At a constrained extremum the gradient of f is parallel to the gradient of the constraint function; B — The solutions of the equations are candidates, so f must be compared at all of them
Tangency of level curve and constraint means parallel gradients, and the method yields only candidate points. λ may have any sign (for x + y = 2 with f = −(x² + y²) it is negative at a maximum). On the whole line xy = x(10 − x) → −∞, so there is a maximum 25 but no minimum.
Evaluate I = ∫₀^∞ x³ e^(−2x) dx and report the value of 8I.
Numerical answer — type the value.
Show answer
Answer: 3
Substituting t = 2x, I = (1/2⁴)∫₀^∞ t³e^(−t) dt = Γ(4)/16 = 6/16 = 3/8, so 8I = 3. Numerical check: Simpson's rule on [0, 30] with 3000 steps gives 0.375000.
The region under y = x² for 0 ≤ x ≤ 2 is rotated about the x-axis, giving a solid of volume V. Find V/π to one decimal place.
Numerical answer — type the value.
Show answer
Answer: 6.4
V = π∫₀² x⁴ dx = π·2⁵/5 = 32π/5, so V/π = 6.4. Python midpoint-sum of x⁴ with 10⁵ strips gives 6.40000.
Maximise f(x, y) = x y² subject to x + y = 9 with x ≥ 0 and y ≥ 0. The maximum value of f is ____.
Numerical answer — type the value.
Show answer
Answer: 108
∇f = λ∇g gives y² = λ and 2xy = λ, so y = 2x (the point y = 0 gives f = 0). With x + y = 9: x = 3, y = 6, f = 3 × 36 = 108. Check: substituting y = 9 − x, f = x(9 − x)², f′ = (9 − x)(9 − 3x) = 0 at x = 3 gives 108, and a grid search over x in [0, 9] gives the same maximum.
The maximum of f(x, y) = x + 2y on the circle x² + y² = 5 is ____.
Numerical answer — type the value.
Show answer
Answer: 5
1 = 2λx, 2 = 2λy give y = 2x, 5x² = 5, so (1, 2) gives 5 and (−1, −2) gives −5. Cauchy-Schwarz check: x + 2y ≤ √(1 + 4)·√(x² + y²) = √5·√5 = 5.
A thin lamina occupies the triangle x ≥ 0, y ≥ 0, x + y ≤ 1 (x, y in metres) with surface density ρ = 12(x + y) kg/m². Its mass in kg is ____.
Numerical answer — type the value.
Show answer
Answer: 4
M = 12∫₀¹∫₀^1−x (x + y) dy dx = 12∫₀¹ [x(1 − x) + (1 − x)²/2] dx = 12(1/6 + 1/6) = 4. By symmetry ∬x dA = ∬y dA = 1/6 (area 1/2, centroid coordinate 1/3), so 12(1/6 + 1/6) = 4 again.
The area enclosed by the cardioid r = a(1 + cos θ), 0 ≤ θ ≤ 2π, using A = ½∫ r² dθ, is:
1. πa²
2. 2πa²
3. 3πa²/2
4. 3πa²/4
Show answer
Answer: C — 3πa²/2
A = ½a²∫₀^2π (1 + 2cos θ + cos² θ) dθ = ½a²(2π + 0 + π) = 3πa²/2. The cos² term contributes π, not 2π; forgetting the ½ in the area formula would give 3πa².