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The following surface integral is to be evaluated over a sphere for the given steady vector field, F = xi + yj + zk defined with respect to a Cartesian coordinate system having i, j, and k as unit base vectors.
, Where S is the sphere, x^{2} + y^{2} + z^{2} = 1 and n is the outward unit normal vector to the sphere. The value of the surface integral is
Concept:
Gauss divergence theorem:
It states that the surface integral of the normal component of a vector function taken over a closed surface ‘S’ is equal to the volume integral of the divergence of that vector function taken over a volume enclosed by the closed surface ‘S’.
Calculation:
Given:
F = xi + yj + zk
Stokes theorem:
It states that the line integral of a vector field around any closed surface C is equal to the surface integral of the normal component of the curl of vector over an unclosed surface ‘S’.
Green's theorem:
The value of the integral , where D is the shaded triangular region shown in the diagram, is _____ (rounded off to the nearest integer).
Calculation:
To cover shaded region
x → 0 to 4
y → x to x
I = 2× 4^{4} = 512
A definite double integral is given below, then, evaluation of the double integral over the region R will be __
Where R is the region on X  Y plane for the function given as, and r ∈ [0, 5]
Concept:
Let’s change the Cartesian coordinate limits into cylindrical planar polar coordinate,
⇒ x = r cosθ; y = r sinθ
Replace dx.dy = rdθ .dr
Calculation:
Given:
θ → 0 to π; r → 0 to 5
Then the limit of the double integration will be,
⇒ I =  [cos π – cos 0] × 625
⇒ I = 1250
Concept:
The triple integral can be evaluated as repeated integral:
First, f(x, y, z) is integrated w.r.t z between limits z_{1} and z_{2} keeping x and y constant, then integrated w.r.t y between limits y_{1} and y_{2} keeping x constant.
The result is then integrated w.r.t x.
Calculation:
Given:
=
Concept:
Triple integral can be evaluated as repeated integral:
First, f(x, y, z) is integrated w.r.t z between limits z_{1} and z_{2} keeping x and y constant, then integrated w.r.t y between limits y_{1} and y_{2} keeping x constant. The result is then integrated w.r.t x.
Calculation:
=
=
=
The integral , where D denotes the disc ��^{2} + ��^{2} ≤ 4, evaluates to__________.
Given,
and x^{2} + y^{2} ≤ 4
Putting x = r.cosθ, y = r.sinθ and dx.dy = r.dr.dθ
The area bounded by the curves y^{2 }= 9x, x – y + 2 = 0 is given by
Calculation
Given equations are: y^{2} = 9x, x – y + 2 = 0
By solving the above two equations,
The point of intersection of the two curves are: (1, 3) and (4, 6)
Now, the graph is shown below.
By considering the horizontal strip,
The limits of y are: 3 to 6
The limits of x are: (y – 2) to y^{2}/9
Now, the required area is =∫∫dxdy
= 1/2
Concept:
Ellipse
Length major axis of ellipse = 2a
Length of minor axis of ellipse = 2b
Area =
Calculation:
For the first quadrant, take a vertical strip as shown. Here, y coordinate varies from 0 to
Also, the xcoordinate varies from 0 to a
∴ Area =
=
=
=
∴ The total area of ellipse = 4 × πab/4 = πab units
We have the integration given as,
Placing the limits we get,
Hence the required value of integration will be 26/105.
Concept:
In a double integral with variable limits, the change of order of integration can sometimes make the evaluation easy.
Calculation:
Since the limit of y is constant hence, we can say that a horizontal strip was taken for integration now for change of order we will take vertical strip as shown in the following diagram.
∴ is the correct solution.
The area between the parabolas y^{2} = 4ax and x^{2} = 4ay is
y^{2} = 4ax & x^{2} = 4ay
We have to find shaded region area.
So area drawn by y= x^{2} / 4a on xaxis = A_{1} (say)
Then shaded area = A_{1} – A_{2}
So,
⇒
⇒ Shaded area =
Let , where R is the region shown in the figure and c = 6 × 10^{4}. The value of I equals________. (Give the answer up to two decimal places.)
= 1
An ellipse is revolved around the yaxis. The volume generated by the solid of revolution if a = 3 and b = 2 is______
Concept:
The volume of solid generated by revolving around yaxis
V = ∫πx^{2}dy (1)
Volume generated by revolving around xaxis
V = ∫πy^{2}dx (2)
Calculations:
From the equation of ellipse
Subtitling in (1)
Given
a = 3
b = 2
V = 43π3^{2}2 = 24π
= 75.39
The value of , where dA indicate small area in xy  plane, is
Concept:
Since the inside limit is in terms of x, therefore we have to integrate first the 'y' terms and convert whole expression in terms of x.
Calculation:
Given:
= 1/3
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