Green's Theorem MCQ Quiz in मल्याळम - Objective Question with Answer for Green's Theorem - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Apr 7, 2025
Latest Green's Theorem MCQ Objective Questions
Top Green's Theorem MCQ Objective Questions
Green's Theorem Question 1:
Evaluate \(\mathop \smallint \limits_c \left[ {\left( {2{x^2} - {y^2}} \right)dx + \left( {{x^2} + {y^2}} \right)dy} \right],\) when C is the boundary of the area enclosed by the x – axis and the upper half of the circle x2 + y2 = a2
Answer (Detailed Solution Below)
Green's Theorem Question 1 Detailed Solution
By Green’s theorem
\(\mathop \smallint \limits_c \left( {\phi dx + \psi dy} \right) = \mathop \int\!\!\!\int \limits_E \left( {\frac{{\partial \psi }}{{\partial x}} - \frac{{\partial \phi }}{{\partial y}}} \right)dx\;dy\)
\(\mathop \smallint \limits_c \left[ {\left( {2{x^2} - {y^2}} \right)dx + \left( {{x^2} + {y^2}} \right)dy} \right] = \mathop \int\!\!\!\int \limits_E {\left[ {\frac{\partial }{{\partial x}}\left( {{x^2} + {y^2}} \right) - \frac{\partial }{{\partial y}}\left( {2{x^2} - {y^2}} \right)} \right]_{dxdy}}\)
\(= 2\mathop \int\!\!\!\int \limits_E \left( {x + y} \right)dxdy\)
Now changing to polar co-ordinates, r varies from 0 to a and Q varies from 0 to π
\(\begin{array}{l} = 2\mathop \smallint \limits_0^a \mathop \smallint \limits_0^\pi r\left( {cos\theta + sin\theta } \right)rd\theta dr\\ = 2\mathop \smallint \limits_0^a {r^2}dr\mathop \smallint \limits_0^\pi \left( {cos\theta + sin\theta } \right)d\theta \\ = 2\mathop \smallint \limits_0^a {r^2}dr.\left( 2 \right) = \left. {4\frac{{{r^3}}}{3}\;} \right]_0^a = \frac{{4{a^3}}}{3} \end{array}\)
Green's Theorem Question 2:
The area of the region enclosed by a simple closed curve C will be _________.
Answer (Detailed Solution Below)
Green's Theorem Question 2 Detailed Solution
Concept:
Green's Theorem:
Green's Theorem relates a line integral around a simple closed curve C to a double integral over the plane region D bounded by C The theorem states:
\(\oint_C (P \, dx + Q \, dy) = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \)
Calculations:
the area enclosed by C, we can use a specific form of Green's Theorem. Consider the vector field (P, Q) =(x/2. y/2). Applying Green's Theorem to this vector field gives us:
\(\oint_C \left( -\frac{y}{2} \, dx + \frac{x}{2} \, dy \right) = \iint_D \left( \frac{\partial}{\partial x} \left( \frac{x}{2} \right) - \frac{\partial}{\partial y} \left( -\frac{y}{2} \right) \right) \, dA \)
Calculating the partial derivatives, we get:
\(\frac{\partial}{\partial x} \left( \frac{x}{2} \right) = \frac{1}{2}\)
\(\frac{\partial}{\partial y} \left( -\frac{y}{2} \right) = -\frac{1}{2}\)
So, the integrand becomes:
\(\iint_D \left( \frac{1}{2} - \left( -\frac{1}{2} \right) \right) \, dA = \iint_D (1) \, dA = \iint_D dA\)
The right-hand side is simply the area of D. Therefore, the left-hand side gives us the area enclosed by C
\(\oint_C \left( -\frac{y}{2} \, dx + \frac{x}{2} \, dy \right) = \text{Area}(D)\)
Multiplying through by 2, we get:
\(\oint_C (-y \, dx + x \, dy) = 2 \cdot \text{Area}(D)\)
Thus, the area A enclosed by C is:
\(A = \frac{1}{2} \oint_C (x \, dy - y \, dx)\)
Conclusion:
Therefore, the correct formula for the area of the region enclosed by a simple closed curve C is:
\(A = \frac{1}{2} \oint_C (x \, dy - y \, dx)\)
Among the given options, the correct answer is: 2
Green's Theorem Question 3:
Evaluate ∮c y3dx - x3dy where c is positively oriented circle of radius 2 centered at origin.
Answer (Detailed Solution Below)
Green's Theorem Question 3 Detailed Solution
Calculations:
In this particular case, it's cleaner to apply Green's Theorem in polar coordinates. The equality of the line integral around the positively oriented circle of radius 2 centered at origin to a double integral over the region D enclosed by the curve is as follows:
∮C P dx + Q dy = ∫∫D (dQ/dx - dP/dy) dA
We first need to compute dQ/dx and dP/dy:
dQ/dx = d(-x3)/dx = -3x2 dP/dy = d(y3)/dy = 3y2
Then, dQ/dx - dP/dy = -3x2 - 3y2
In polar coordinates, x = rcos(θ) and y = rsin(θ), and the area element dA in polar coordinates is r dr dθ.
Replacing x and y with these and simplifying gives:
dQ/dx - dP/dy = -3r2(cos2(θ) + sin2(θ)) = -3r2.
So, by Green's Theorem, the line integral ∮C P dx + Q dy is equal to the double integral
∫ (from 0 to 2π) ∫ (from 0 to 2) -3r2 × r dr dθ.
Compute this double integral to obtain the final result. Let's do this:
= ∫ (from 0 to 2π) [(-3/4 × r4) from 0 to 2] dθ = ∫ (from 0 to 2π) (-3/4 × 16) dθ = -12 × ∫ (from 0 to 2π) dθ = -12 × [θ from 0 to 2π] = -12 × 2π = -24π.
The value of ∮C y3 dx - x3 dy around the given circle in the positive direction is -24π.
Green's Theorem Question 4:
Pick the correct statement?
Answer (Detailed Solution Below)
Green's Theorem Question 4 Detailed Solution
Let C be a positively oriented, piecewise smooth, simple closed curve in a plane,
and let D be the region bounded by C.
If L and M are functions of (x, y) defined on an open region containing D and have continuous partial derivatives there, then
where the path of integration along C is anticlockwise.
Stoke's Theorem -
Let be a smooth oriented surface in with boundary . If a vector field is defined and has continuous first order partial derivatives in a region containing , then
Here, Green's Theorem is a Particular form of Stoke's Theorem .
Therefore, Correct Option is Option 4).
Green's Theorem Question 5:
Which theorem is valid in calculating integral for a multiple connected domain R?
Answer (Detailed Solution Below)
Green's Theorem Question 5 Detailed Solution
Explanation:
Let C be the positively oriented, smooth, and simple closed curve in a plane, and D be the region bounded by the C. If P and Q are the functions of (x, y) defined on the open region, containing D and have continuous partial derivatives, then Green’s theorem is stated as
\(\int_C (Pdx+Qdy)\) = \(\int\int_D\left({\partial Q\over \partial x}-{\partial P\over \partial y}\right)dxdy\)
Therefore, Green's theorem is valid in calculating integral for a multiple connected domain R.
Option (1) is true.
Green's Theorem Question 6:
The value of \(\oint\limits_{\rm{C}} {{\rm{[(cos}}} \,{\rm{x}}\,{\rm{sin}}\,{\rm{y}}\,{\rm{ - xy)dx + sin}}\,{\rm{x}}\,{\rm{cos}}\,{\rm{y}}\,{\rm{dy]}}\) where C is the circle x2 + y2 = 1, is:
Answer (Detailed Solution Below)
Green's Theorem Question 6 Detailed Solution
Concept:
Green's Theorem:- If two functions M(x, y) and N(x, y and their partial derivatives are single valued and continuous over a region R bounded by a closed curve C, then
\(\oint \left(M dx + Ndy \right) = \int \int_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dx dy\)
Green Theorem is useful for evaluating a line integral around a closed curve C.
Calculation:
We have,
⇒ \(\oint\limits_{\rm{C}} {{\rm{[(cos}}} \,{\rm{x}}\,{\rm{sin}}\,{\rm{y}}\,{\rm{ - xy)dx + sin}}\,{\rm{x}}\,{\rm{cos}}\,{\rm{y}}\,{\rm{dy]}}\)
On comparing, we get
⇒ M = cos x sin y - x y
⇒ N = sin x cos y
On differentiating M partially with respect to 'y'
⇒ \(\frac{\partial M}{\partial y} = cos x \ cosy - x\)
On differentiating N partially with respect to 'x'
⇒ \(\frac{\partial N}{\partial x} = cos x \ cosy\)
⇒ \(\int \int_s [(cos x \ cos y) - (cos x cos y - x) ]dx dy\)
⇒ \(\int \int_s x dx dy\)
⇒ \(\displaystyle \int_{{x = - 1}}^1 \int_ { y = - \sqrt{1 - x^2}}^{\sqrt{1 - x^2}}\ \ \ x dx dy\)
⇒ \(\displaystyle \int_{-1}^1 x \displaystyle \left[y \right]_{- \sqrt{1 - x^2}}^{\sqrt{1 - x^2}}dx \)
⇒ \(\displaystyle \int_{-1}^{1} x dx \left[ \sqrt{1 - x^2} + \sqrt{1 - x^2} \right] \)
⇒ \(2 \displaystyle \int_{-1}^1 x \sqrt{1 - x^2} dx\)
⇒ Let 1 - x2 = t
⇒ On differentiating, we get
⇒ - 2x dx = dt
⇒ x dx = \(- \frac{dt}{2}\)
When x = -1
⇒ 1 - 1 = 0 = t
When x = 1
⇒ 1 - 1 = 0 = t
⇒ \(2 \displaystyle \int_{-0}^0 - \sqrt{t} \frac{dt}{2}\)
⇒ 2 × 0
⇒ 0
∴ The value of \(\oint\limits_{\rm{C}} {{\rm{[(cos}}} \,{\rm{x}}\,{\rm{sin}}\,{\rm{y}}\,{\rm{ - xy)dx + sin}}\,{\rm{x}}\,{\rm{cos}}\,{\rm{y}}\,{\rm{dy]}}\) where C is the circle x2 + y2 = 1 is 0.
Green's Theorem Question 7:
The value of ∫ xy dy – y2dx, where C is rectangle cut from the first quadrant by the lines x = 1 and y = 2 is
Answer (Detailed Solution Below)
Green's Theorem Question 7 Detailed Solution
Concept:
By Green’s theorem
\(I = \mathop \smallint \nolimits_C Mdx + Ndy = \mathop \int\!\!\!\int \limits_A^{} \left( {\frac{{\partial N}}{{\partial x}} - \frac{{\partial M}}{{\partial y}}} \right)dx\;dy{{\;\;\;\;\;}} \ldots \left( 1 \right)\)
Calculation:
Given:
M = xy, N = – y2
\(\Rightarrow \frac{{\partial N}}{{\partial x}} - \frac{{\partial M}}{{\partial y}} = y - \left( { - 2y} \right) = 3y\)
By using equation (1),
\(\Rightarrow I = \smallint xy\;dy - {y^2}dx = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^2 \left( {3y} \right)dy\;dx\)
\(\Rightarrow I = \mathop \smallint \limits_{x = 0}^1 \frac{3}{2}\left[ {{y^2}} \right]_0^2dx\)
\(\Rightarrow I = \mathop \smallint \limits_0^1 6\;dx\)
\(\therefore \mathop \smallint \nolimits_C xy\;dy - {y^2}dx = 6\)
Green's Theorem Question 8:
The value of the integral \(\mathop \smallint \limits_c^{} \left[ {\left( {2{x^2} - {y^2}} \right)dx + \left( {{x^2} + {y^2}} \right)dy} \right]\) where C is the boundary of the area enclosed by the axis and the upper half of the circle x2 + y2 = a2 . Take a = 3.
Answer (Detailed Solution Below) 72
Green's Theorem Question 8 Detailed Solution
By using Green’s Theorem use have
\(\mathop \oint \limits_c^{} \left( {Pdx + Qdy} \right) = \mathop \int\!\!\!\int \limits_s^{} \left[ {\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}}} \right]dxdy\)
\(\mathop \smallint \limits_c^{} \left( {2{x^2} - {y^2}} \right)dx + \left( {{x^2} + {y^2}} \right)dy \)
\(= \mathop \smallint \limits_{-a}^a \mathop \smallint \limits_0^{\sqrt {{a^2} - {x^2}} } \left[ {\frac{\partial }{{\partial x}}\left( {{x^2} + {y^2}} \right) - \frac{\partial }{{\partial y}}\left( {2{x^2} - {y^2}} \right)} \right]dxdy\)
\(= \mathop \smallint \limits_{-a}^a \mathop \smallint \limits_0^{\sqrt {{a^2} - {x^2}} } \left[ {2x + 2y} \right]dxdy\)
\(= 2\mathop \smallint \limits_{ - a}^a \mathop \smallint \limits_0^{\sqrt {{a^2} - {x^2}} } \left( {x + y} \right)dxdy\)
\(= 2\mathop \smallint \limits_{-a}^a dx\mathop \smallint \limits_0^{\sqrt {{a^2} - {x^2}} } \left( {xy + \frac{{{y^2}}}{2}} \right)dy\)
\(= 2\mathop \smallint \limits_{-a}^a dx\left[ {x\sqrt {{a^2} - {x^2}} + \frac{{{a^2} - {x^2}}}{2}} \right]\)
\(= 2\mathop \smallint \limits_{-a}^a \left( {x\sqrt {{a^2} - {x^2}} } \right)dx + \mathop \smallint \limits_{ - a}^a \left( {{a^2} - {x^2}} \right)dx\)
\(\mathop \smallint \limits_0^a f\left( x \right)dx = = \left\{ {\begin{array}{*{20}{c}} {2\mathop \smallint \limits_0^a f\left( x \right)dx,\;\;if\;f\;is\;even}\\ {0,\;\;if\;f\;is\;odd} \end{array}} \right.\)
\(= 2[0 + 2\mathop \smallint \limits_0^a \left( {{a^2} - {x^2}} \right)dx]\)
\(= 4\left[ {{a^2}x - \frac{{{x^3}}}{3}} \right]_0^a\)
\(= 4\left[ {{a^3} - \frac{{{a^3}}}{3}} \right]\)
\({l} = 4 \times \frac{{2{a^3}}}{3} = \frac{{8{a^3}}}{3} \)
With a = 3, we get the integrated value as
l = 72
Green's Theorem Question 9:
Find the value of \(\mathop \oint \limits_c^\; \left[ {\left( {xy + {y^2}} \right)dx + {x^2}dy} \right],\) where C is bounded by y = x and y = x2.
Answer (Detailed Solution Below)
Green's Theorem Question 9 Detailed Solution
According to Green’s theorem,
\(\mathop \smallint \limits_c^\; \left( {M dx + N dy} \right) = \mathop \int\!\!\!\int \limits_E^\; \left( {\frac{{\partial {\rm{N }}}}{{\partial {\rm{x}}}} - \frac{{\partial M }}{{\partial y}}} \right)dx\;dy\)
Here in the given question,
M = xy + y2
\(\frac{{\partial M }}{{\partial y}} = x + 2y\)
N = x2
\(\frac{{\partial N }}{{\partial x}} = 2x\)
\(\mathop \oint \limits_C^\; \left[ {\left( {xy + {y^2}} \right)dx + {x^2}dy} \right] = \mathop \int\!\!\!\int \limits_E^\; \left( {2x - x - 2y} \right)dx\;dy = \mathop \int\!\!\!\int \limits_E^\; \left( {x - 2y} \right)dx\;dy\)
\(\begin{array}{l} \mathop \int\!\!\!\int \limits_E^\; \left( {x - 2y} \right)dx\;dy\\ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = {x^2}}^x \left( {x - 2y} \right)dx\;dy\\ = \mathop \smallint \limits_0^1 \left[ {xy - {y^2}} \right]_{{x^2}}^xdx\\ = \mathop \smallint \limits_0^1 \left[ {{x^4} - {x^3}} \right]dx = \left[ {\frac{{{x^5}}}{5} - \frac{{{x^4}}}{4}} \right]_0^1 = - \frac{1}{{20}} \end{array}\)
Green's Theorem Question 10:
Green's theorem says that the total "microscopic circulation" in D is equal to the circulation around the boundary c = dD̅
Answer (Detailed Solution Below)
Green's Theorem Question 10 Detailed Solution
Explanation:
Green's theorem states that the circulation form of the double integral over a plane region D is equal to the line integral around the boundary curve C of D (often noted as ∂D). To put it quantitatively:
∫∫_D (curl F) × da = ∮_C F × ds
Where:
The left side is a double integral over the region D of the curl of a vector field F dotted with da, the area element. This represents the total "microscopic circulation" in D.
The right side is a line integral around the closed curve C (the boundary of D, ∂D) of F dotted with ds, the line element. This represents the circulation around the boundary C.