BMTC-102 Solved Assignment January 2026 | बहुचर कलन | IGNOU BSCG

Q: State whether the following statements are true or false. Give reasons for your answers.

Answer available — download the full PDF for complete details.

Q1. State whether the following statements are true or false. Give reasons for your answers.

(i)) lim (x->0) x^2 sin(1/x) / sin x = 1 (80 words)
(ii)) A real-valued function of three variables which is continuous everywhere is differentiable. (80 words)
(iii)) The function F : R^2 → R^2, defined by F(x, y) = (y + 2, x + y), is locally invertible at any (x, y) ∈ R^2. (80 words)
(iv)) f: [-1,1]×[−2,2] → R, defined by f(x, y) = {x, if y is rational; 0, if y is not rational} is integrable. (80 words)
(v)) The function f : R^2 → R, defined by f(x, y) = 1 − y^2 + x^2, has an extremum at (0,0). (80 words)

Key Points:

L'Hopital's Rule with Squeeze Theorem for evaluating limits like `x sin(1/x)`.
Continuity is necessary but not sufficient for differentiability in multivariable functions.
Inverse Function Theorem: Local invertibility requires a non-zero Jacobian determinant.
Riemann Integrability requires the function's set of discontinuities to have measure zero.

Answer:

Q3(a). Let the function f be defined by f(x, y) = {3x^2 y^4 / (x^8 + y^4), (x, y) ≠ (0,0); 0, (x, y) = (0,0)}. Show that f has directional derivatives in all directions at (0,0).

Key Points:

Directional derivative at (a,b) is D_u f(a,b) = lim (h→0) [f(a+hu₁, b+hu₂) - f(a,b)] / h.
For (0,0), substitute x=hu₁, y=hu₂ into f(x,y) for (x,y) ≠ (0,0).
f(hu₁, hu₂) simplifies to [3 h² u₁² u₂⁴] / [h⁴ u₁⁸ + u₂⁴] for h ≠ 0.
The expression for D_u f(0,0) simplifies to lim (h→0) [3 h u₁² u₂⁴] / [h⁴ u₁⁸ + u₂⁴].

Answer: The directional derivative of a function f(x,y) at a point (a,b) in the direction of a unit vector u = <u₁, u₂> is defined as D_u f(a,b) = lim (h→0) [f(a + h u₁, b + h u₂) - f(a,b)] / h. For f(x, y) at (0,0), with f(0,0) = 0, the formula becomes D_u f(0,0) = lim (h→0) [f(h u₁, h u₂) - 0] / h. Substituting x = h u₁ and y = h u₂ into the function definition for (x,y) ≠ (0,0): f(h u₁, h u₂) = [3(h u₁)²(h u₂)⁴] / [(h u₁)⁸ + (h u₂)⁴] = [3 h² u₁² h⁴ u₂⁴] / [h⁸ u₁⁸ + h⁴ u₂⁴] = [3 h⁶ u₁² u₂⁴] / [h⁴(h⁴ ...

Q3(b). Let x = e^r cos θ, y = e^r sin θ and f be a continuously differentiable function of x and y, whose partial derivatives are also continuously differentiable. Show that ∂^2 f / ∂r^2 + ∂^2 f / ∂θ^2 = (x^2 + y^2) (∂^2 f / ∂x^2 + ∂^2 f / ∂y^2).

Key Points:

Coordinate transformation given: x = e^r cos θ, y = e^r sin θ.
Calculated first partial derivatives of x, y with respect to r, θ.
Applied chain rule to find ∂f/∂r and ∂f/∂θ expressions.
Derived ∂²f/∂r² using product rule and multivariable chain rule.

Answer: The problem requires demonstrating the equality ∂²f/∂r² + ∂²f/∂θ² = (x² + y²) (∂²f/∂x² + ∂²f/∂y²) given the coordinate transformation x = e^r cos θ and y = e^r sin θ. First, we compute the necessary partial derivatives of x and y with respect to r and θ, which are fundamental for the chain rule application: ∂x/∂r = e^r cos θ = x ∂y/∂r = e^r sin θ = y ∂x/∂θ = -e^r sin θ = -y ∂y/∂θ = e^r cos θ = x Using the chain rule, the first partial derivatives of f (a continuously differentiable function) w...

Q4(a). Find the centre of gravity of a thin sheet with density δ(x, y) = y, bounded by the curves y = 4x^2 and x = 4.

Key Points:

Center of gravity (x̄, ȳ) is found using mass (M) and moments (M_x, M_y).
Mass M = ∫∫_R δ(x, y) dA is the total mass of the sheet.
Moments M_x = ∫∫_R y δ(x, y) dA and M_y = ∫∫_R x δ(x, y) dA are calculated.
The region R is 0 ≤ x ≤ 4 and 0 ≤ y ≤ 4x², assuming common boundary conditions.

Answer: To find the center of gravity (x̄, ȳ) of a thin sheet, we first need to calculate its total mass (M) and its moments about the x and y axes (M_x and M_y, respectively). The density function is given as δ(x, y) = y, and the region is bounded by the curves y = 4x² and x = 4. We interpret this as the region in the first quadrant, bounded by y = 4x², the x-axis (y=0), and the line x = 4. Therefore, the region of integration R is defined by 0 ≤ x ≤ 4 and 0 ≤ y ≤ 4x². First, we calculate the total m...

Q4(b). Find the mass of the solid bounded by z = 1 and z = x^2 + y^2, the density function being δ(x, y, z) = |x|.

Key Points:

Mass calculation involves a triple integral M = ∫∫∫_E δ dV.
Solid defined by z = x² + y² and z = 1, projecting to unit disk x² + y² ≤ 1.
Inner integral dz: ∫_(x²+y²)^1 |x| dz = |x|(1 - x² - y²).
Polar coordinates (r, θ) simplify integration for circular regions.

Answer: To find the mass of the solid, we first define the region of integration. The solid is bounded below by the paraboloid z = x² + y² and above by the plane z = 1. These surfaces intersect when x² + y² = 1, forming a unit circle in the xy-plane. This circular region, D, will be the projection of the solid onto the xy-plane. The mass M is calculated using a triple integral of the density function δ(x, y, z) over the solid E: M = ∫∫∫_E δ(x, y, z) dV. Given δ(x, y, z) = |x|, the integral becomes M =...

Q5(a). State Green's theorem, and apply it to evaluate ∫_C [(3x^2-4y) dx - (2x + y^3) dy, Where C is the ellipse 4x^2 +9y^2 = 36.

Key Points:

Green's Theorem converts line integrals into double integrals.
Formula: ∫_C (P dx + Q dy) = ∫∫_D (∂Q/∂x - ∂P/∂y) dA.
Identify P and Q from the line integral components.
Calculate partial derivatives: ∂P/∂y and ∂Q/∂x.

Answer: Green's Theorem states that if C is a positively oriented, piecewise smooth, simple closed curve in the plane and D is the region bounded by C, and if P and Q are functions with continuous first partial derivatives on an open region containing D, then: ∫_C (P dx + Q dy) = ∫∫_D (∂Q/∂x - ∂P/∂y) dA. For the given integral, ∫_C [(3x²-4y) dx - (2x + y³) dy], we identify P = (3x²-4y) and Q = -(2x + y³) = -2x - y³. Next, we calculate the required partial derivatives: ∂P/∂y = ∂/∂y (3x²-4y) = -4, and ∂...

Q5(b). Find the extreme values of the function f(x, y) = x^2 + y on the surface x^2 + 2y^2 = 1.

Key Points:

Lagrange Multipliers method is used for constrained optimization problems.
Define objective function f(x, y) and constraint g(x, y) = 0.
Set up gradient equations: ∂f/∂x = λ(∂g/∂x) and ∂f/∂y = λ(∂g/∂y).
Solve the system of equations derived from ∇f = λ∇g and g(x,y)=0.

Answer: To find the extreme values of f(x, y) = x² + y on the surface x² + 2y² = 1, we use the method of Lagrange Multipliers, a key technique for constrained optimization taught in BMTC-102. We define the objective function f(x, y) = x² + y and the constraint g(x, y) = x² + 2y² - 1 = 0. The Lagrange Multiplier conditions are ∇f = λ∇g and the constraint g(x,y)=0. The partial derivatives are: f_x = 2x, f_y = 1, g_x = 2x, g_y = 4y. This gives the system: 1) 2x = λ(2x) 2) 1 = λ(4y) 3) x² + 2y² = 1 From ...

Q6(a). Check the continuity and differentiability of the function at (0,0) where f(x, y) = {2x^3 y / (x^4 + y^2), (x, y) ≠ (0,0); 0, otherwise}

Key Points:

Continuity requires lim f(x,y) = f(a,b) at the point (a,b).
f(0,0) is given as 0 for this piecewise function.
The limit for continuity at (0,0) is evaluated as 0 using polar coordinates.
Partial derivatives fₓ(0,0) and fᵧ(0,0) are both 0, calculated by definition.

Answer: To check the continuity and differentiability of the function f(x, y) at (0,0), we apply the definitions from Multivariable Calculus (BMTC-102). First, for continuity, a function f(x,y) is continuous at (a,b) if lim (x,y)→(a,b) f(x,y) = f(a,b). Here, f(0,0) = 0. We evaluate the limit lim (x,y)→(0,0) [2x³y / (x⁴ + y²)]. Using polar coordinates, let x = r cosθ and y = r sinθ. The limit becomes: lim (r→0) [2(r cosθ)³(r sinθ) / ((r cosθ)⁴ + (r sinθ)²)] = lim (r→0) [2r⁴ cos³θ sinθ / (r⁴ cos⁴θ + r² ...

Q6(b). Find the domain and range of the function f, defined by f(x, y) = 2xy / (x^2 + y^2). Also find two level curves of this function. Give a rough sketch of them.

Key Points:

Domain of f(x,y) = 2xy/(x²+y²) is all R² except the origin (0,0).
Range of f(x,y) = 2xy/(x²+y²) is the closed interval [-1, 1].
Range is derived using inequalities: -1 ≤ 2xy/(x²+y²) ≤ 1.
A level curve for f(x,y) is defined by f(x,y) = k (constant).

Answer: The function is defined as f(x, y) = 2xy / (x² + y²). **Domain:** For f(x, y) to be defined, the denominator cannot be zero. Thus, x² + y² ≠ 0. This condition is violated only when both x=0 and y=0. Therefore, the domain of f is all points in the xy-plane except the origin (0, 0), denoted as R² \ {(0, 0)}. **Range:** To determine the range, consider the algebraic identity (x - y)² ≥ 0, which implies x² - 2xy + y² ≥ 0, or x² + y² ≥ 2xy. Since x² + y² > 0 for points in the domain, we can divide ...

Q7(a). Evaluate ∫_C (2x^2 + 3y^2) dx, where C is the curve given by x(t) = at^2, y(t) = 2at, 0 ≤ t ≤ 1.

Key Points:

Line integrals ∫_C P dx are evaluated by parameterizing the curve.
Substitute x=x(t), y=y(t), and dx = x'(t) dt into the integral.
The integration limits change to the parameter's range, typically [t₁, t₂].
For x(t)=at², dx/dt = 2at, leading to dx = 2at dt.

Answer: To evaluate the line integral ∫_C (2x² + 3y²) dx, we parameterize the curve C and express the integral in terms of the parameter t. This method is discussed in Unit 3, Section 3.2.1 of BMTC-102, which covers line integrals of real-valued functions. The given curve is defined by x(t) = at² and y(t) = 2at, for 0 ≤ t ≤ 1. First, we need to find dx/dt. Differentiating x(t) with respect to t, we get dx/dt = d/dt (at²) = 2at. Therefore, dx = 2at dt. Next, substitute x(t) and y(t) into the integrand ...

Q7(b). Use double integration of find the volume of the ellipsoid x^2/4 + y^2/9 + z^2/16 = 1.

Key Points:

Volume by double integration is V = ∫∫_R 2z dA for symmetric solids.
Ellipsoid equation: x²/a² + y²/b² + z²/c² = 1, where a=2, b=3, c=4.
Projection onto xy-plane is an ellipse R: x²/4 + y²/9 = 1.
Change of variables (x=2u, y=3v) transforms R to unit circle R' (u²+v²=1).

Answer: To find the volume of the ellipsoid x²/4 + y²/9 + z²/16 = 1 using double integration, we first express the volume V as a double integral. The volume of a solid bounded by a surface z = f(x,y) and the xy-plane over a region R is given by V = ∫∫_R f(x,y) dA. Since the ellipsoid is symmetric about the xy-plane, we can calculate the volume of the upper half (z ≥ 0) and multiply by two. From the ellipsoid equation, we solve for z: z² = 16(1 - x²/4 - y²/9), so z = 4√(1 - x²/4 - y²/9) for the upper ha...

Q8(a). Find the values of a and b, if lim (x->∞) (x(1+a cos x)-b sin x) / x^3 = 1.

Key Points:

Use Maclaurin series for `cos x` (`1 - x²/2! + ...`) and `sin x` (`x - x³/3! + ...`).
Substitute series into the numerator: `x(1+a cos x) - b sin x`.
Collect terms by powers of `x`: `(1+a-b)x + (-a/2 + b/6)x³ + ...`.
For limit to be finite, coefficient of `x` term must be zero: `1 + a - b = 0`.

Answer: To find the values of `a` and `b`, we analyze the given limit expression. This type of problem, involving trigonometric functions and a specific finite non-zero limit with a polynomial denominator, typically requires the use of Taylor series expansions, especially when coefficients need to be determined. Although the limit is for `x -> ∞`, such problems are often implicitly solved by considering the Taylor expansion around `x=0` to identify terms that must vanish or equate to the limit value. L...

Q8(b). Suppose S and C are subsets of R^3. S is the unit open sphere with centre at the origin and C is the open cube= {P(x, y, z) | − 1 < x < 1,−1 < y < 1,−1 < z <1}. Which of the following is true. Justify your answer. (i) S ⊂ C (ii) C ⊂ S

Key Points:

Unit open sphere S: points (x,y,z) with x² + y² + z² < 1.
Open cube C: points (x,y,z) with -1 < x < 1, -1 < y < 1, -1 < z < 1.
For S ⊂ C: If x² + y² + z² < 1, then each x², y², z² < 1, implying -1 < x,y,z < 1.
Thus, every point in S is also in C, making S ⊂ C true.

Answer: To determine which statement is true, S ⊂ C or C ⊂ S, we must understand the definitions of the unit open sphere (S) and the open cube (C) in R³ and analyze their geometric properties. The unit open sphere S, centered at the origin, consists of all points P(x, y, z) in R³ such that their distance from the origin is less than 1. Mathematically, S = { (x, y, z) ∈ R³ | x² + y² + z² < 1 }. This definition aligns with the concept of an open ball in Rⁿ discussed in Unit 1, Section 1.3 of BMTC-102. T...

Q8(c). Identify the level curves of the following functions:

(i)) sqrt(x^2 + y^2) (40 words)
(ii)) sqrt(4 - x^2 - y^2) (40 words)

Key Points:

Level curves represent points (x,y) where f(x,y) equals a constant `k`.
For `f(x,y) = sqrt(x^2 + y^2)`, level curves are concentric circles `x^2 + y^2 = k^2`.
These circles are centered at the origin (0,0) with radius `k`, for `k ≥ 0`.
For `f(x,y) = sqrt(4 - x^2 - y^2)`, level curves are concentric circles `x^2 + y^2 = 4 - k^2`.

Answer: To identify level curves, we set the function f(x,y) equal to a constant `k`. (i) For `f(x,y) = sqrt(x^2 + y^2)`, setting `f(x,y) = k` gives `sqrt(x^2 + y^2) = k`. Squaring both sides yields `x^2 + y^2 = k^2`. These are concentric circles centered at the origin (0,0) with radius `k`, where `k ≥ 0`. (ii) For `f(x,y) = sqrt(4 - x^2 - y^2)`, setting `f(x,y) = k` yields `sqrt(4 - x^2 - y^2) = k`. Squaring both sides gives `x^2 + y^2 = 4 - k^2`. These are concentric circles centered at the origin (...

Q9(a). Using polar coordinates, show that lim (x->0, y->0) (x^3 - y^3) / (x^2 + y^2) = 0. Also, find the two repeated limits.

Key Points:

Substitute x=r cos(θ), y=r sin(θ) for polar coordinate limit.
As (x, y) → (0, 0), r → 0 in polar coordinates.
Simplify (x³ - y³) / (x² + y²) to r(cos³(θ) - sin³(θ)).
Limit becomes 0 because r → 0 and (cos³(θ) - sin³(θ)) is bounded.

Answer: To evaluate the limit using polar coordinates, we substitute x = r cos(θ) and y = r sin(θ) into the given expression. As (x, y) approaches (0, 0), the radial distance r approaches 0. Substituting these into the expression f(x, y) = (x³ - y³) / (x² + y²), we get: f(r cos(θ), r sin(θ)) = ( (r cos(θ))³ - (r sin(θ))³ ) / ( (r cos(θ))² + (r sin(θ))² ) = ( r³ cos³(θ) - r³ sin³(θ) ) / ( r² cos²(θ) + r² sin²(θ) ) = r³ (cos³(θ) - sin³(θ)) / ( r² (cos²(θ) + sin²(θ)) ) = r³ (cos³(θ) - sin³(θ)) / ( r² × 1...

Q9(b). Write ∫_0^1 ∫_0^sqrt(1-x^2) dy dx as an integral over a region D. Sketch the region D and show that it is of both types 1 and 2. Reverse the order of integration and evaluate it.

Key Points:

Integral limits define the region D of integration.
The region D is a quarter-circle of radius 1 in the first quadrant.
A Type 1 region has y bounded by functions of x for constant x range.
A Type 2 region has x bounded by functions of y for constant y range.

Answer: The given integral is ∫_0^1 ∫_0^sqrt(1-x^2) dy dx. The inner limits for y are from 0 to sqrt(1-x²), and the outer limits for x are from 0 to 1. This integral defines the region D. The upper limit y=sqrt(1-x²) implies x² + y² = 1 for y ≥ 0. Combined with x from 0 to 1, D is the quarter-circle of radius 1 in the first quadrant, bounded by x=0, y=0, and the arc x² + y² = 1. To sketch D, draw the x and y axes. Mark the origin, (1,0) on the x-axis, and (0,1) on the y-axis. Draw a circular arc conne...

Q10(a). Check if the following integrals are independent of path and evaluate those which are independent.

(i)) ∫_(0,0)^(3,4) (6xy − y^3) dx + (3x^2 – x^2y) dy (100 words)
(ii)) ∫_(-1,4)^(3,8) [(3x^2 - 2y^2) dx-4xy dy] (100 words)

Key Points:

Line integral path independence: ∂P/∂y = ∂Q/∂x condition.
If ∂P/∂y = ∂Q/∂x, the vector field is conservative.
Conservative fields allow evaluation using a potential function f(x,y).
Fundamental Theorem of Line Integrals: ∫ F · dr = f(B) - f(A).

Answer: To determine if the given line integrals are independent of path, we employ the condition for a conservative vector field, which states that ∂P/∂y must equal ∂Q/∂x. If this condition is satisfied, the integral is path-independent and can be efficiently evaluated using a potential function f(x, y), applying the Fundamental Theorem of Line Integrals.

Q10(b). Evaluate ∫∫∫_S z^2 dx dy dz, where S is the solid region between the spheres ρ=1 and ρ=2, by using spherical coordinates.

Key Points:

Spherical coordinates simplify triple integrals over spherical regions.
Conversion formulas: `z = ρ cos φ`, `dx dy dz = ρ² sin φ dρ dφ dθ`.
Region between spheres `ρ=1` and `ρ=2` implies `1 ≤ ρ ≤ 2`, `0 ≤ φ ≤ π`, `0 ≤ θ ≤ 2π`.
The integral becomes `∫_0^(2π) ∫_0^π ∫_1^2 ρ⁴ cos² φ sin φ dρ dφ dθ`.

Answer: To evaluate the triple integral ∫∫∫_S z^2 dx dy dz, we convert to spherical coordinates as detailed in Unit 4, Section 4.5.3 of BMTC-102. The `z` coordinate transforms to `z = ρ cos φ`. The volume element `dx dy dz` transforms to `ρ² sin φ dρ dφ dθ`, representing the Jacobian for spherical coordinates. Substituting these into the integral, the integrand `z²` becomes `(ρ cos φ)²`. The integral then takes the form `∫∫∫ ρ⁴ cos² φ sin φ dρ dφ dθ`. For the solid region between spheres `ρ=1` and `ρ=2...

BMTC-102 Solved Assignment — January 2026 / July 2026

बहुचर कलन | IGNOU Bachelor of Science (General) (CBCS) (BSCG)

BMTC-102—January 2026 / July 2026

बहुचर कलन

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BMTC-102 Assignment Questions — January 2026

Q1. State whether the following statements are true or false. Give reasons for your answers.

Q3(a). Let the function f be defined by f(x, y) = {3x^2 y^4 / (x^8 + y^4), (x, y) ≠ (0,0); 0, (x, y) = (0,0)}. Show that f has directional derivatives in all directions at (0,0).

Q3(b). Let x = e^r cos θ, y = e^r sin θ and f be a continuously differentiable function of x and y, whose partial derivatives are also continuously differentiable. Show that ∂^2 f / ∂r^2 + ∂^2 f / ∂θ^2 = (x^2 + y^2) (∂^2 f / ∂x^2 + ∂^2 f / ∂y^2).

Q4(a). Find the centre of gravity of a thin sheet with density δ(x, y) = y, bounded by the curves y = 4x^2 and x = 4.

Q4(b). Find the mass of the solid bounded by z = 1 and z = x^2 + y^2, the density function being δ(x, y, z) = |x|.

Q5(a). State Green's theorem, and apply it to evaluate ∫_C [(3x^2-4y) dx - (2x + y^3) dy, Where C is the ellipse 4x^2 +9y^2 = 36.

Q5(b). Find the extreme values of the function f(x, y) = x^2 + y on the surface x^2 + 2y^2 = 1.

Q6(a). Check the continuity and differentiability of the function at (0,0) where f(x, y) = {2x^3 y / (x^4 + y^2), (x, y) ≠ (0,0); 0, otherwise}

Q6(b). Find the domain and range of the function f, defined by f(x, y) = 2xy / (x^2 + y^2). Also find two level curves of this function. Give a rough sketch of them.

Q7(a). Evaluate ∫_C (2x^2 + 3y^2) dx, where C is the curve given by x(t) = at^2, y(t) = 2at, 0 ≤ t ≤ 1.

Q7(b). Use double integration of find the volume of the ellipsoid x^2/4 + y^2/9 + z^2/16 = 1.

Q8(a). Find the values of a and b, if lim (x->∞) (x(1+a cos x)-b sin x) / x^3 = 1.

Q8(b). Suppose S and C are subsets of R^3. S is the unit open sphere with centre at the origin and C is the open cube= {P(x, y, z) | − 1 < x < 1,−1 < y < 1,−1 < z <1}. Which of the following is true. Justify your answer. (i) S ⊂ C (ii) C ⊂ S

Q8(c). Identify the level curves of the following functions:

Q9(a). Using polar coordinates, show that lim (x->0, y->0) (x^3 - y^3) / (x^2 + y^2) = 0. Also, find the two repeated limits.

Q9(b). Write ∫_0^1 ∫_0^sqrt(1-x^2) dy dx as an integral over a region D. Sketch the region D and show that it is of both types 1 and 2. Reverse the order of integration and evaluate it.

Q10(a). Check if the following integrals are independent of path and evaluate those which are independent.

Q10(b). Evaluate ∫∫∫_S z^2 dx dy dz, where S is the solid region between the spheres ρ=1 and ρ=2, by using spherical coordinates.

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