When a function depends on two or more variables, a partial derivative measures how it changes with one variable while the others are kept constant. This topic covers partial derivatives, Euler's theorem on homogeneous functions, total derivatives and chain rule, Jacobians, and maxima–minima of functions of two variables.
Key points
For u = f(x, y): ∂u/∂x is found by treating y as a constant; ∂u/∂y by treating x as constant.
Homogeneous function of degree n: f(tx, ty) = tⁿ f(x, y).
Euler's theorem: if u is homogeneous of degree n, x ∂u/∂x + y ∂u/∂y = nu.
Corollary: x²uₓₓ + 2xy uₓᵧ + y²uᵧᵧ = n(n − 1)u.
If f(u) is homogeneous of degree n (u itself not), then x uₓ + y uᵧ = n f(u)/f'(u).
Total derivative and chain rule: if u = f(x, y) with x, y functions of t, du/dt = (∂u/∂x)(dx/dt) + (∂u/∂y)(dy/dt).
Jacobian J = ∂(u, v)/∂(x, y) measures how areas change under a change of variables.
Maxima and minima: at a stationary point fₓ = 0, fᵧ = 0. With r = fₓₓ, s = fₓᵧ, t = fᵧᵧ: rt − s² > 0 and r < 0 → maximum; rt − s² > 0 and r > 0 → minimum; rt − s² < 0 → saddle point; rt − s² = 0 → needs more study.
Lagrange's multipliers handle extremum with a condition (constraint).
For Euler's theorem questions, first find the degree carefully by replacing x by tx and y by ty. If u is inside sin⁻¹, tan⁻¹ or log, apply Euler to the inner homogeneous function.
Maxima and Minima of Two Variables and Lagrange's Multipliers
Quick summary
A function f(x, y) of two variables can have maximum, minimum and saddle points. We find stationary points using partial derivatives and classify them with the second-derivative test (rt − s²). When the variables must satisfy a condition, Lagrange's method of undetermined multipliers is used. Taylor's theorem for two variables gives the basis for the test.
Key points
Taylor's theorem (two variables): f(a + h, b + k) = f(a, b) + (h fₓ + k fᵧ) + (1/2!)(h² fₓₓ + 2hk fₓᵧ + k² fᵧᵧ) + …, all derivatives at (a, b).
Stationary point: where fₓ = 0 and fᵧ = 0 together.
Notation at a stationary point: r = fₓₓ, s = fₓᵧ, t = fᵧᵧ.
Test: if rt − s² > 0 and r < 0 → maximum; rt − s² > 0 and r > 0 → minimum; rt − s² < 0 → saddle point (neither); rt − s² = 0 → doubtful, needs further study.
An extreme value is a local (relative) value, not necessarily the largest or smallest over the whole plane.
Lagrange's multipliers: to find extreme values of f(x, y, z) subject to φ(x, y, z) = 0, form F = f + λφ and solve ∂F/∂x = 0, ∂F/∂y = 0, ∂F/∂z = 0 together with φ = 0.
Lagrange's method gives the stationary points; the nature (max or min) is usually decided from the physical meaning of the problem.
Important terms
Saddle point – a stationary point which is a max along one direction and a min along another.
Extreme value (कमाल-किमान मूल्य) – a maximum or minimum value.
Lagrange multiplier λ – an unknown constant introduced to include the condition.
Formulas
Stationary point: fₓ = 0, fᵧ = 0
r = fₓₓ, s = fₓᵧ, t = fᵧᵧ; decide by the sign of rt − s² and of r
F = f + λφ; Fₓ = Fᵧ = F_z = 0, φ = 0
5 important questions
Q1. What is a saddle point?
A stationary point (fₓ = fᵧ = 0) at which rt − s² < 0. The function increases in some directions and decreases in others, so it is neither a maximum nor a minimum. Example: (0, 0) for f = x² − y².
Q2. State the conditions for f(x, y) to have a maximum at (a, b).
fₓ = 0 and fᵧ = 0 at (a, b), and at that point rt − s² > 0 with r < 0 (where r = fₓₓ, s = fₓᵧ, t = fᵧᵧ).
Q3. Find the extreme values of f(x, y) = x³ + y³ − 3axy (a > 0).
fₓ = 3x² − 3ay = 0 and fᵧ = 3y² − 3ax = 0, so y = x²/a and x = y²/a.
Substituting: x = x⁴/a³, so x(x³ − a³) = 0, giving x = 0 or x = a. Points: (0, 0) and (a, a).
r = 6x, s = −3a, t = 6y.
At (0, 0): rt − s² = 0 − 9a² < 0 → saddle point.
At (a, a): r = 6a, t = 6a, rt − s² = 36a² − 9a² = 27a² > 0 and r > 0 → minimum.
Minimum value = a³ + a³ − 3a·a·a = −a³.
Q4. Find the minimum of x² + y² + z² subject to x + y + z = 6.
F = x² + y² + z² + λ(x + y + z − 6).
Fₓ = 2x + λ = 0, Fᵧ = 2y + λ = 0, F_z = 2z + λ = 0, so x = y = z = −λ/2.
From the condition: 3x = 6, so x = y = z = 2.
Minimum value = 4 + 4 + 4 = 12. (It is a minimum, since x² + y² + z² can be made as large as we like on the plane.)
Q5. Find the maximum of f(x, y) = 2x + 4y − x² − y² + 5.
fₓ = 2 − 2x = 0 → x = 1; fᵧ = 4 − 2y = 0 → y = 2.
r = −2, s = 0, t = −2; rt − s² = 4 > 0 and r < 0 → maximum.
Maximum value = 2 + 8 − 1 − 4 + 5 = 10.
Exam tip
Make a small table of each stationary point with r, s, t, rt − s² and the conclusion. In Lagrange problems, first express all variables in terms of λ, then use the condition.
A reduction formula connects an integral containing a power n with a similar integral of lower power (n − 1 or n − 2). Using it again and again, we reduce the power until the integral is easy. The most used results are for sinⁿx, cosⁿx, sinᵐx cosⁿx and tanⁿx, especially between 0 and π/2 (Wallis' formula).
Key points
Reduction formulae are obtained by integration by parts, writing sinⁿx = sinⁿ⁻¹x · sin x and so on.
Wallis' formula for sinᵐx cosⁿx (0 to π/2): numerator = (m − 1)(m − 3)… × (n − 1)(n − 3)…, denominator = (m + n)(m + n − 2)…; multiply by π/2 only when m and n are both even.
Each product of factors stops at 1 or 2.
Important terms
Reduction formula (लघुकरण सूत्र) – a formula expressing Iₙ through Iₙ₋₁ or Iₙ₋₂.
Wallis' formula – direct value of ∫₀^(π/2) sinⁿx dx or sinᵐx cosⁿx dx.
Use Iₙ = 1/(n − 1) − Iₙ₋₂ with I₀ = ∫₀^(π/4) dx = π/4.
I₂ = 1/1 − I₀ = 1 − π/4.
I₄ = 1/3 − I₂ = 1/3 − 1 + π/4 = π/4 − 2/3.
Check: π/4 − 2/3 ≈ 0.785 − 0.667 = 0.118, positive and small, as expected since 0 ≤ tan⁴x ≤ 1 there.
Exam tip
For Wallis' formula, write the numerator and denominator factors in two clear lines and decide "× π/2 or not" before simplifying. Derivations are asked often — learn the integration-by-parts step for sinⁿx well.
Beta and Gamma functions are special functions defined by improper integrals. Many hard definite integrals can be written in terms of them and evaluated quickly. The key facts are Γ(n + 1) = nΓ(n), Γ(½) = √π and the relation B(m, n) = Γ(m)Γ(n)/Γ(m + n).
A double integral adds up a function over a region of the plane; a triple integral does the same over a solid. They give areas, volumes, mass and centres of gravity. We learn to evaluate them as repeated integrals, change the order of integration, and change variables (polar, cylindrical, spherical) using the Jacobian.
Key points
∬_R f(x, y) dx dy is evaluated as a repeated integral: integrate with respect to one variable (treating the other as constant), then the other.
When the inner limits depend on the outer variable, the inner integral must be done first.
Area of R = ∬_R dx dy; volume under z = f(x, y) = ∬_R f dx dy; volume of a solid V = ∭ dx dy dz.
Change of order: draw the region, then describe it again with the other variable outside. This often turns an impossible inner integral into an easy one.
Change of variables: dx dy = |J| du dv where J = ∂(x, y)/∂(u, v).
Polar: x = r cos θ, y = r sin θ, dx dy = r dr dθ. Useful for circles and for x² + y².
Cylindrical: dx dy dz = r dr dθ dz. Spherical: dx dy dz = r² sin θ dr dθ dφ.
Order of limits in a triple integral: innermost first, from variable limits to constant limits.
Important terms
Region of integration – the area (or solid) over which we integrate.
Jacobian (याकोबियन) – the scale factor between areas (or volumes) in old and new variables.
Repeated (iterated) integral – a double integral written as two single integrals.
Formulas
Area = ∬ dx dy = ∬ r dr dθ
Mass = ∬ ρ dx dy; x̄ = (∬ xρ dx dy)/mass
∫₀^∞ e^(−x²) dx = √π/2 (proved using polar coordinates)
5 important questions
Q1. Evaluate ∫₀¹ ∫₀ˣ (x + y) dy dx.
Inner: ∫₀ˣ (x + y) dy = xy + y²/2 from 0 to x = x² + x²/2 = 3x²/2. Outer: ∫₀¹ 3x²/2 dx = ½. Answer: ½.
Q2. Find the area of a circle of radius a using a double integral.
In polar coordinates: Area = ∫₀^(2π) ∫₀ᵃ r dr dθ = ∫₀^(2π) (a²/2) dθ = (a²/2)(2π) = πa².
Q3. Change the order and evaluate ∫₀¹ ∫ₓ¹ e^(y²) dy dx.
The inner integral ∫ e^(y²) dy has no elementary form, so change the order.
Region: 0 ≤ x ≤ 1, x ≤ y ≤ 1, i.e. the triangle with vertices (0,0), (0,1), (1,1).
Described the other way: 0 ≤ y ≤ 1, 0 ≤ x ≤ y.
Integral = ∫₀¹ ∫₀^y e^(y²) dx dy = ∫₀¹ y e^(y²) dy.
Put y² = t: = ½ ∫₀¹ eᵗ dt = (e − 1)/2.
Q4. Evaluate ∫₀^∞ ∫₀^∞ e^(−(x² + y²)) dx dy and deduce ∫₀^∞ e^(−x²) dx.
The region is the first quadrant: 0 ≤ r < ∞, 0 ≤ θ ≤ π/2.
Integral = ∫₀^(π/2) ∫₀^∞ e^(−r²) r dr dθ.
∫₀^∞ e^(−r²) r dr = ½ (put r² = t). So the integral = (π/2)(½) = π/4.
But the integral also equals (∫₀^∞ e^(−x²) dx)² = I². So I² = π/4 and I = √π/2.
Q5. Find the volume of the tetrahedron bounded by x = 0, y = 0, z = 0 and x + y + z = 1.
V = ∫₀¹ ∫₀^(1−x) ∫₀^(1−x−y) dz dy dx.
Inner: ∫ dz = 1 − x − y.
Middle: ∫₀^(1−x) (1 − x − y) dy = (1 − x)² − (1 − x)²/2 = (1 − x)²/2.
Outer: ∫₀¹ (1 − x)²/2 dx = ½ × ⅓ = 1/6.
Volume = 1/6 cubic units.
Exam tip
Always sketch the region before changing the order of integration — most mistakes come from wrong limits. In polar problems, never forget the extra factor r.
A differential equation relates a function with its derivatives. This topic covers first-order, first-degree equations and the standard methods to solve them: variables separable, homogeneous, linear, Bernoulli's and exact equations. Simple applications such as orthogonal trajectories and growth–decay problems are also included.
Key points
Order = highest derivative present; degree = power of the highest derivative (after removing radicals and fractions).
The general solution has as many arbitrary constants as the order; a particular solution fixes them using given conditions.
Variables separable: f(x) dx = g(y) dy → integrate both sides.
Homogeneous equation: dy/dx = f(y/x). Put y = vx, dy/dx = v + x dv/dx, then separate.
Linear equation: dy/dx + Py = Q (P, Q functions of x). Integrating factor IF = e^(∫P dx). Solution: y × IF = ∫ Q × IF dx + c.
Bernoulli's equation: dy/dx + Py = Qyⁿ. Divide by yⁿ and put z = y^(1−n); it becomes linear in z.
Exact equation: M dx + N dy = 0 is exact if ∂M/∂y = ∂N/∂x.
Solution of exact equation: ∫M dx (y constant) + ∫(terms of N not containing x) dy = c.
If not exact, an integrating factor may make it exact.
Orthogonal trajectories: curves cutting a given family at right angles; replace dy/dx by −dx/dy in the family's equation.
Q1. Find the order and degree of (d²y/dx²)³ + (dy/dx)⁴ + y = 0.
The highest derivative is d²y/dx², so the order is 2. Its power is 3, so the degree is 3.
Q2. Solve dy/dx = xy.
Separate: dy/y = x dx. Integrate: log y = x²/2 + c. So y = A e^(x²/2), where A is a constant.
Q3. Solve dy/dx + y = eˣ.
Linear with P = 1, Q = eˣ.
IF = e^(∫1 dx) = eˣ.
y eˣ = ∫eˣ · eˣ dx + c = e^(2x)/2 + c.
y = eˣ/2 + c e^(−x).
Check: y' = eˣ/2 − c e^(−x); y' + y = eˣ ✓.
Q4. Solve (2xy + 3) dx + (x² + 4y) dy = 0.
M = 2xy + 3, N = x² + 4y.
∂M/∂y = 2x, ∂N/∂x = 2x. Equal, so the equation is exact.
∫M dx (y constant) = x²y + 3x.
Terms of N without x: 4y; ∫4y dy = 2y².
Solution: x²y + 3x + 2y² = c.
Q5. Solve the homogeneous equation dy/dx = (x + y)/x.
Put y = vx, dy/dx = v + x dv/dx.
v + x dv/dx = (x + vx)/x = 1 + v.
x dv/dx = 1 → dv = dx/x.
v = log x + c → y/x = log x + c.
Solution: y = x log x + cx.
Exam tip
First identify the type (separable, homogeneous, linear, Bernoulli, exact) — write it as the first line of your answer. For linear equations, always write P, Q and IF clearly.
First Order Higher Degree Equations and Clairaut's Form
Quick summary
A first-order differential equation of higher degree contains powers of p = dy/dx, such as p² or p³. Such equations are solved by factorising in p (solvable for p), or by making them solvable for y or for x. Clairaut's equation y = px + f(p) is the most important special case: it has a general solution (a family of straight lines) and a singular solution (their envelope).
Key points
Notation: p = dy/dx. The equation has the form F(x, y, p) = 0 of degree n in p.
Solvable for p: factorise into (p − f₁)(p − f₂)…(p − fₙ) = 0, solve each first-order, first-degree equation, and write the general solution as the product of all the solutions = 0, using one constant c.
Solvable for y: write y = f(x, p), differentiate with respect to x to get an equation in x and p, solve it, then eliminate p (or keep p as a parameter).
Solvable for x: write x = f(y, p), differentiate with respect to y using dx/dy = 1/p.
Clairaut's equation: y = px + f(p). The general solution is obtained by simply replacing p by c: y = cx + f(c).
Singular solution of Clairaut's equation: eliminate p between y = px + f(p) and x + f′(p) = 0. It contains no arbitrary constant and is not a special case of the general solution.
Geometrically, the singular solution is the envelope of the family of lines y = cx + f(c).
Some equations reduce to Clairaut's form by a substitution such as X = x², Y = y².
Important terms
Degree of a DE – the power of the highest derivative after making the equation free of radicals and fractions.
Singular solution (विचित्र उकल) – a solution with no arbitrary constant that cannot be got from the general solution.
Envelope – a curve touching every member of a family of curves.
Formulas
Solvable for y: y = f(x, p) ⇒ p = ∂f/∂x + (∂f/∂p)(dp/dx)
Clairaut: y = px + f(p) ⇒ general y = cx + f(c); singular: x + f′(p) = 0 with the given equation
5 important questions
Q1. Solve p² − 5p + 6 = 0.
(p − 2)(p − 3) = 0, so dy/dx = 2 or 3, giving y = 2x + c or y = 3x + c. General solution: (y − 2x − c)(y − 3x − c) = 0.
Q2. What is Clairaut's equation? Write its general solution.
An equation of the form y = px + f(p), where p = dy/dx. Its general solution is y = cx + f(c), found by replacing p with the constant c.
Q3. Solve y = px + a/p and find its singular solution.
This is Clairaut's form with f(p) = a/p. General solution: y = cx + a/c.
For the singular solution, differentiate the right side with respect to p: x − a/p² = 0, so p² = a/x and p = √(a/x).
p + y = −x is linear: dy/dx + y = −x, IF = eˣ. So y eˣ = −∫x eˣ dx = −(x − 1)eˣ + c, giving y = 1 − x + c e^(−x).
General solution: (2y − x² − c)(y + x − 1 − c e^(−x)) = 0.
Exam tip
When you see y = px + f(p), write the general solution in one line and spend your effort on the singular solution. In "solvable for p" problems, check your factorisation by expanding before solving.
When three vectors are multiplied, we get either a scalar (scalar triple product) or a vector (vector triple product). The scalar triple product gives the volume of a parallelepiped and tests whether vectors are coplanar. The vector triple product has a useful expansion formula. Products of four vectors and reciprocal vectors are built from these.
Key points
Scalar triple product: [a b c] = a · (b × c). For a = a₁i + a₂j + a₃k etc., it equals the 3 × 3 determinant with rows (a₁, a₂, a₃), (b₁, b₂, b₃), (c₁, c₂, c₃).
Cyclic order does not change it: [a b c] = [b c a] = [c a b]; swapping two vectors changes the sign. Dot and cross can be interchanged: a · (b × c) = (a × b) · c.
|[a b c]| = volume of the parallelepiped with edges a, b, c. Volume of the tetrahedron = (1/6)|[a b c]|.
Three vectors are coplanar iff [a b c] = 0. If two of the three vectors are equal, the product is 0.
Vector triple product: a × (b × c) = (a · c)b − (a · b)c. It lies in the plane of b and c.
In general a × (b × c) ≠ (a × b) × c, so the cross product is not associative.
Four vectors: (a × b) · (c × d) = (a · c)(b · d) − (a · d)(b · c) (Lagrange's identity); (a × b) × (c × d) = [a b d]c − [a b c]d.
Reciprocal vectors: a′ = (b × c)/[a b c], b′ = (c × a)/[a b c], c′ = (a × b)/[a b c]; then a · a′ = 1 and a · b′ = 0.
Important terms
Box product – another name for [a b c].
Coplanar vectors (समतलीय सदिश) – vectors lying in the same plane.
Parallelepiped – a solid with six parallelogram faces.
Formulas
[a b c] = det[a₁ a₂ a₃; b₁ b₂ b₃; c₁ c₂ c₃]
a × (b × c) = (a · c)b − (a · b)c
[a + b, b + c, c + a] = 2[a b c]
5 important questions
Q1. Show that [a b c] = 0 if a, b, c are coplanar.
b × c is perpendicular to the plane of b and c. If a lies in the same plane, a is perpendicular to b × c, so a · (b × c) = 0.
Q2. State the formula for the vector triple product.
a × (b × c) = (a · c)b − (a · b)c. Memory aid: "outer dot far, times near, minus outer dot near, times far".
Q3. Find the volume of the parallelepiped with edges a = 2i − 3j + 4k, b = i + 2j − k, c = 3i − j + 2k.
Write the determinant neatly and expand along the first row with signs + − +. For coplanarity questions, set the determinant to zero; for volume, take the absolute value.
Vector differentiation deals with vectors that change with a variable (such as time) and with fields that vary in space. The del operator ∇ gives three key ideas: gradient of a scalar field, and divergence and curl of a vector field. These are basic tools in physics (fluid flow, electricity, magnetism).
Key points
If r(t) = x(t)i + y(t)j + z(t)k, then dr/dt = (dx/dt)i + (dy/dt)j + (dz/dt)k. For a moving particle, velocity v = dr/dt, acceleration a = d²r/dt².
A vector of constant magnitude is perpendicular to its derivative.
Del operator: ∇ = i ∂/∂x + j ∂/∂y + k ∂/∂z.
Gradient (प्रवणता) of scalar φ: ∇φ, a vector normal to the level surface φ = c, pointing in the direction of fastest increase.
Directional derivative of φ along unit vector â is ∇φ · â. Its maximum value is |∇φ|.
Divergence of F: ∇·F, a scalar; measures the net outflow per unit volume. If ∇·F = 0, F is solenoidal.
Curl of F: ∇ × F, a vector; measures rotation. If ∇ × F = 0, F is irrotational (and F = ∇φ for some scalar potential φ).
Important identities: curl(grad φ) = 0; div(curl F) = 0.
∇·r = 3 and ∇ × r = 0 for the position vector r = xi + yj + zk; ∇r = r/r (r = |r|).
Laplacian: ∇²φ = φₓₓ + φᵧᵧ + φ_zz.
Important terms
Scalar field – a number at each point (temperature).
Vector field – a vector at each point (velocity of a fluid).
Level surface – surface on which φ has a constant value.
Q1. Define solenoidal and irrotational vector fields.
F is solenoidal if div F = 0 everywhere (no sources or sinks). F is irrotational if curl F = 0 everywhere; then F can be written as the gradient of a scalar potential.
Q2. Find the unit normal to the surface x² + y² + z² = 9 at (2, 1, 2).
φ = x² + y² + z²; ∇φ = 2xi + 2yj + 2zk = 4i + 2j + 4k at the point. |∇φ| = √(16 + 4 + 16) = 6. Unit normal = (2i + j + 2k)/3.
Q3. Find the directional derivative of φ = x²y + yz² at (1, 1, 1) in the direction of i + 2j + 2k.
Similarly the j and k components are φ_zx − φₓ_z = 0 and φᵧₓ − φₓᵧ = 0.
(Mixed partial derivatives are equal for smooth φ.)
Hence curl(grad φ) = 0.
Exam tip
Write curl as a 3 × 3 determinant with i, j, k in the first row and ∂/∂x, ∂/∂y, ∂/∂z in the second — it avoids sign mistakes. Always reduce the direction to a unit vector first.
Vector integration extends integration to curves, surfaces and volumes. Line integrals give work done by a force; surface integrals give flux through a surface. Three great theorems — Green's, Stokes' and Gauss's divergence theorem — connect these integrals to each other.
Key points
Line integral ∫_C F·dr = ∫(F₁ dx + F₂ dy + F₃ dz) along curve C. If F is a force, it gives the work done.
If F is conservative (F = ∇φ, curl F = 0), the line integral depends only on the end points, and around a closed curve it is zero.
Surface integral ∬_S F·n̂ dS gives the flux of F across surface S.
Volume integral ∭_V φ dV.
Green's theorem (plane): relates a line integral round a closed curve C to a double integral over the region R it encloses.
Stokes' theorem: the line integral of F round the boundary of an open surface equals the surface integral of curl F over that surface.
Gauss's divergence theorem: the outward flux of F through a closed surface equals the volume integral of div F over the volume inside.
Green's theorem is the plane (2-D) case of Stokes' theorem.
Curve C is taken anticlockwise (positive direction) in Green's theorem.
Important terms
Closed curve – starts and ends at the same point (∮).
Flux (अभिवाह) – amount of field passing through a surface.
Outward normal – unit normal pointing out of a closed surface.
Simply connected region – region without holes.
Formulas
Work W = ∫_C F·dr
Green: ∮_C (M dx + N dy) = ∬_R (∂N/∂x − ∂M/∂y) dx dy
Area by Green: A = ½∮(x dy − y dx)
Stokes: ∮_C F·dr = ∬_S (∇ × F)·n̂ dS
Gauss: ∯_S F·n̂ dS = ∭_V (∇·F) dV
5 important questions
Q1. State Gauss's divergence theorem.
If V is a volume bounded by a closed surface S and F has continuous first derivatives, then ∯_S F·n̂ dS = ∭_V div F dV, where n̂ is the outward unit normal.
Q2. When is a line integral independent of path?
When F is conservative, i.e. F = ∇φ for some scalar φ (equivalently curl F = 0 in a simply connected region). Then ∫ from A to B of F·dr = φ(B) − φ(A).
Q3. Evaluate ∫_C F·dr for F = x² i + xy j along the line y = x from (0, 0) to (1, 1).
F·dr = x² dx + xy dy.
On y = x: dy = dx, xy = x².
Integral = ∫ from 0 to 1 of (x² + x²) dx = ∫₀¹ 2x² dx.
= 2[x³/3]₀¹ = 2/3.
Q4. Using Green's theorem, evaluate ∮_C (−y dx + x dy) where C is the circle x² + y² = a².
M = −y, N = x.
∂N/∂x − ∂M/∂y = 1 − (−1) = 2.
∮ = ∬_R 2 dx dy = 2 × (area of circle).
= 2πa².
(This also shows area = ½∮(x dy − y dx) = πa².)
Q5. Using the divergence theorem, find the flux of F = x i + y j + z k over the sphere x² + y² + z² = a².
div F = 1 + 1 + 1 = 3.
Flux = ∭_V 3 dV = 3 × volume of sphere.
= 3 × (4/3)πa³ = 4πa³.
(Direct check: on the sphere F·n̂ = a, so flux = a × 4πa² = 4πa³ ✓.)
Exam tip
For "verify" questions, do both sides separately and show they are equal. For "evaluate using the theorem", convert to the easier side — usually the double or volume integral.