Solved Numericals for Gauss, Stokes and Green's Theorem

Solved Numericals

Q1: Let C be the curve of intersection of the cylinder x² + y² = 4 and the plane z - 2 = 0. Suppose C is oriented in the counterclockwise direction around the 𝑧-axis, when viewed from above. If

then the value of 𝛼 equals _______.
Ans: 16
C is the curve of intersection of the cylinder x² + y² = 4 and the plane z - 2 = 0
So, S :  x² + y² = 4
Also,  
Hence

So, curl

Hence = $<img\; src="https://cn.edurev.in/ApplicationImages/Temp/1611942\_751c468f-0f81-4254-8cdd-41907cd43be2\_lg.png"\; style="max-width:calc(100\%\; -\; 5px);word-break:break-word;width:auto;display:block;margin:5px\; auto;text-align:center;letter-spacing:inherit\; !important"\; fr-original-class="fr-draggable"\; loading="lazy"\; data-gumlet="false"\; alt="Solved\; Numericals">$
Now, S is a circle of radius 4
So, putting x = 2cosθ, y = 2sinθ we get

Therefore απ = 16π while implies α = 16.

 
Q2. By applying Stokes theorem, the value of  where C is the boundary of the triangle with vertices (2, 0, 0), (0, 3, 0) and (0, 0, 6), is
Solution: By stokes theorem:

Calculation:
Given:

Equation of the plane through A, B, C is

∴ ϕ = 3x + 2y + z - 6 represents equation of the surface.
Vector normal to the surface is given by gradient:

A (2, 0, 0), B (0, 3, 0) and C (0, 0, 6)

Q3. If C is the circle  x² + y² = 1 taken in anti-clockwise direction then ∫_c[(x²⁰¹⁵ y²⁰¹⁶ + 2014y) dy + (x²⁰¹⁶ y²⁰¹⁵ + 2017x) dy] will be
Solution: Green's theorem: Let C be a positively oriented, piecewise smooth, simple closed curve in the plane and let D be a region bounded by C. If P(x,y) and Q(x,y) have continuous first partial derivative on an open region that contains D, Then 

∫c[(x2015 y2016 + 2014y) dy + (x2016 y2015 + 2017x) dy]

Use green's theorem

∫ ∫_c (2016 x²⁰¹⁵ y²⁰¹⁵+ 2017 - 2016 x2015 y2015 -2014) dx dy

= 3 ∫ ∫_c ds 

= 3 x area of circle 

= 3π 

Q4: The area of the region enclosed by a simple closed curve C will be _________.
Solution: Green's Theorem:
Let C be the positively oriented, smooth, and simple closed curve in a plane, and S be the region bounded by the C. If M and N are the functions of (x, y) defined on the open region containing S, then: 

Using Green's theorem,

Putting M = y and and N = - x, we get:

∴ The area of the region enclosed by a simple closed curve C will be .

Q5. The value of  where C is the circle x² + y² = 1, is: 
Solution: Green's Theorem:
Let C be the positively oriented, smooth, and simple closed curve in a plane, and S be the region bounded by the C. If M and N are the functions of (x, y) defined on the open region containing S, 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
When x = -1
⇒ 1 - 1 = 0 = t
When x = 1
⇒ 1 - 1 = 0 = t

⇒ 2 × 0⇒ 0

Q6. Evaluate ∮c y³dx - x³dy where c is positively oriented circle of radius 2 centered at origin.
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.

Q7. As shown in the figure, 𝐶 is the arc from the point (3, 0) to the point (0, 3) on the circle x² + y² = 9. The value of the integral  is _____ (up to 2 decimal places). 

Ans: 0

According to Green's theorem

Here, M = y² + 2yx
N = 2xy + x²

Q8. The line integral of function  in the counter clock wise direction, along the circle x$\mathrm{<sup\; style="word-break:break-word;line-height:1.8;letter-spacing:inherit\; !important;font-size:13px\; !important"><strong\; style="word-break:break-word;line-height:1.8;letter-spacing:inherit\; !important">2</strong></sup>}$$+$$\mathrm{<strong\; style="word-break:break-word;line-height:1.8;letter-spacing:inherit\; !important">y</strong>}$²$<strong\; style="word-break:break-word;line-height:1.8;letter-spacing:inherit\; !important">=</strong>\mathrm{<strong\; style="word-break:break-word;line-height:1.8;letter-spacing:inherit\; !important">1</strong>}$ at z$<strong\; style="word-break:break-word;line-height:1.8;letter-spacing:inherit\; !important">=</strong>1$ is 
Solution: 
Green's theorem:

$\oint (Mdx+Ndy)=\int \int (\frac{𝜕}{N}-\frac{𝜕}{M})dx$Calculation:

From Green's Theorem

= - Area of circle (x² + y² = 1)
= - π × (1)²
= - π

Q9. If over the path shown in the figure is 

Solution:
Green's Theorem
If "R" is a closed region of xy plane bounded by  1 or more simple curves 'C' and M(x, y), N(x, y) are continuous functions in a region "R" then we have

Given vector is:

Using Green's theorem

Q10. Green's theorem is used to-
Solution: Green's theorem

• It converts the line integral to a double integral. 
• It transforms the line integral in xy - plane to a surface integral on the same xy - plane.

If M and N are functions of (x, y) defined in an open region then from Green's theorem