Green's Theorem MCQ Quiz in தமிழ் - Objective Question with Answer for Green's Theorem - இலவச PDF ஐப் பதிவிறக்கவும்
Last updated on Apr 4, 2025
Latest Green's Theorem MCQ Objective Questions
Top Green's Theorem MCQ Objective Questions
Green's Theorem Question 1:
Evaluate
Answer (Detailed Solution Below)
Green's Theorem Question 1 Detailed Solution
By Green’s theorem
Now changing to polar co-ordinates, r varies from 0 to a and Q varies from 0 to π
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:
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:
Calculating the partial derivatives, we get:
So, the integrand becomes:
The right-hand side is simply the area of D. Therefore, the left-hand side gives us the area enclosed by C
Multiplying through by 2, we get:
Thus, the area A enclosed by C is:
Conclusion:
Therefore, the correct formula for the area of the region enclosed by a simple closed curve C is:
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
Green's Theorem -
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 Σ" id="MathJax-Element-25-Frame" role="presentation" style="position: relative;" tabindex="0">
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
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
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
Green Theorem is useful for evaluating a line integral around a closed curve C.
Calculation:
We have,
⇒
On comparing, we get
⇒ M = cos x sin y - x y
⇒ N = sin x cos y
On differentiating M partially with respect to 'y'
⇒
On differentiating N partially with respect to 'x'
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒ Let 1 - x2 = t
⇒ On differentiating, we get
⇒ - 2x dx = dt
⇒ x dx =
When x = -1
⇒ 1 - 1 = 0 = t
When x = 1
⇒ 1 - 1 = 0 = t
⇒
⇒ 2 × 0
⇒ 0
∴ The value of
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
Calculation:
Given:
M = xy, N = – y2
By using equation (1),
Green's Theorem Question 8:
The value of the integral
Answer (Detailed Solution Below) 72
Green's Theorem Question 8 Detailed Solution
By using Green’s Theorem use have
With a = 3, we get the integrated value as
l = 72
Green's Theorem Question 9:
Find the value of
Answer (Detailed Solution Below)
Green's Theorem Question 9 Detailed Solution
According to Green’s theorem,
Here in the given question,
M = xy + y2
N = x2
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.