Stokes and Gauss Divergence Theorem Questions and Answers

This set of Vector Integral Calculus Multiple Choice Questions & Answers (MCQs) focuses on “Stokes and Gauss Divergence Theorem”.

1. Which of the following is obtained by evaluating ∬_S\(\overrightarrow {r}.\hat{n}\) dS, where S is a closed surface and V is the Volume using Gauss Divergence Theorem?
a) 3V
b) 2V
c) V
d) 5V
View Answer

Answer: a
Explanation:
Gauss Divergence Theorem is
∬_S\(\overrightarrow {F}.\hat{n}\) dS = ∭_V∇.\(\overrightarrow {F}\)dV
∬_S\(\overrightarrow {r}.\hat{n}\) dS = ∭_V∇.\(\overrightarrow {r}\)dV
= ∭_V\(\bigg ( \overrightarrow {i} \frac {∂}{∂x} + \overrightarrow {j} \frac {∂}{∂y} + \overrightarrow {k} \frac {∂}{∂z} \bigg ) \).(x\(\overrightarrow {i}\) + y\(\overrightarrow {j}\) + z\(\overrightarrow {k} \))dV
= ∭_V\(\bigg [ \frac {∂}{∂x}\)(x) + \(\frac {∂}{∂y}\)(y) + \(\frac {∂}{∂z}(z)\bigg ]\)dV
= ∭_V(1 + 1 + 1) dV = 3∭_VdV
∬_S\(\overrightarrow {r}.\hat{n}\) dS = 3V

2. Which of the following is obtained by evaluating ∬_S∇r.\(\hat{n}\) dS, where S is a closed surface and V is the Volume using Gauss Divergence Theorem?
a) 6V
b) 3V
c) 2V
d) V
View Answer

Answer: a
Explanation:
Gauss Divergence Theorem is
∬_S\(\overrightarrow {F}.\hat{n}\) dS = ∭_V∇.\(\overrightarrow {F}\)dV
Here, \(\overrightarrow {F}\) = ∇r², \(\overrightarrow {r}\) = x\(\overrightarrow {i}\) + y\(\overrightarrow {j}\) + \(\overrightarrow {k}\) and r = |\(\overrightarrow {r}\)|
∬_S∇r².\(\hat{n}\) dS = ∭_V∇.(∇r²)dV = ∭_V∇²r²dV
= ∭_V\(\bigg (\frac {∂^2}{∂x^2} + \frac {∂^2}{∂y^2} + \frac {∂^2}{∂x^2} \bigg )\) r² dV
= ∭_V\(\bigg (\frac {∂^2}{∂x^2} + \frac {∂^2}{∂y^2} + \frac {∂^2}{∂x^2} \bigg )\)(x² + y² + z²)dV
Since,r = |\(\overrightarrow {r}\)| = \(\sqrt {x^2 + y^2 + z^2}\)
r² = x² + y² + z²
= ∭_V(2 + 2 + 2)dV = 6∭_VdV
∬_S∇r².\(\hat{n}dS\) = 6V, where V is a volume of the closed surface S.

3. Which of the following is obtained by evaluating ∬_S\(\frac {\overrightarrow {r}.\hat{n}}{r^2}\)dS, where S is a closed surface and V is the Volume using Gauss Divergence Theorem?
a) ∭_V\(\frac {1}{r^2}\)dV
b) ∭_V\(\frac {2}{r^2}\)dV
c) ∭_V\(\frac {3}{r^2}\)dV
d) ∭_V\(\frac {4}{r^2}\)dV
View Answer

Answer: a
Explanation:
We know that
∬_S\(\frac {\overrightarrow {r}.\hat{n}}{r^2}\)dS = ∭_V∇.\(\frac {\overrightarrow {r}}{r^2} \)dV
Now, ∇. \(\frac {\overrightarrow {r}}{r^2} = \bigg (\overrightarrow {i} \frac {∂}{∂x} + \overrightarrow {j} \frac {∂}{∂y} + \overrightarrow {k} \frac {∂}{∂z} \bigg).\bigg (\frac {x\overrightarrow {i} + y\overrightarrow {j} + z\overrightarrow {k}}{x^2 + y^2 + z^2}\bigg )\)
= \(\frac {∂}{∂x} \bigg ( \frac {x}{x^2 + y^2 + z^2} \bigg ) + \frac {∂}{∂y} \bigg ( \frac {y}{x^2 + y^2 + z^2} \bigg ) + \frac {∂}{∂z} \bigg ( \frac {z}{x^2 + y^2 + z^2} \bigg ) \)
= \(\frac {(x^2 + y^2 + z^2) – 2x^2}{(x^2 + y^2 + z^2)^2} + \frac {(x^2 + y^2 + z^2) – 2y^2}{(x^2 + y^2 + z^2)^2} + \frac {(x^2 + y^2 + z^2) – 2z^2}{(x^2 + y^2 + z^2)^2}\)
= \(\frac {1}{x^2 + y^2 + z^2} = \frac {1}{r^2}\)
∬_S\(\frac {\overrightarrow {r}.\hat{n}}{r^2}\)dS = ∭_V∇.\(\frac {1}{r^2} \)dV

4. Which of the following is obtained by evaluating ∬_S\(\overrightarrow {F}.\hat{n}\) dS, where S is a closed surface and V is the Volume using Gauss Divergence Theorem if \(\overrightarrow {F}\) = ax\(\overrightarrow {i}\) + by\(\overrightarrow {j}\) + cz\(\overrightarrow {k}\)?
a) \(\frac {4π}{3}\)(a + b + c)
b) \(\frac {4π}{5}\)(a + b + c)
c) \(\frac {2π}{3}\)(a + b + c)
d) \(\frac {π}{3}\)(a + b + c)
View Answer

Answer: a
Explanation:
Gauss Divergence Theorem is
∬_S\(\overrightarrow {F}.\hat{n}\) dS = ∭_V∇.\(\overrightarrow {F}\)dV
∭_V∇.\(\overrightarrow {F}\)dV = ∭_V\(\bigg ( \frac {∂}{∂x}\) (ax) + \(\frac {∂}{∂y}\) (by) + \(\frac {∂}{∂z}\) (cz)\( \bigg )\)dV
= ∭_V(a + b + c) dV = (a + b + c) ∭_VdV
= \(\frac {4π}{3}\)(a + b + c)

5. Which of the following is obtained by evaluating ∬_S[xz²dydx + (x²y – z³)dzdx + (2xy + y²z)dxdy, where S is a closed surface and V is the Volume using Gauss Divergence Theorem?
a) \(\frac {2}{3}\)πa⁵
b) πa⁵
c) \(\frac {1}{3}\)πa⁵
d) \(\frac {2}{5}\)πa⁵
View Answer

Answer: a
Explanation:
We know that, Cartesian form of Gauss’s divergence theorem is
∬_S[xz²dydx + (x²y – z³)dzdx + (2xy + y²z)dxdy = ∭_V [\(\frac {∂}{∂x}\) (xz²) + \(\frac {∂}{∂y}\)(x²y – z³) +
\(\frac {∂}{∂z}\)(2xy + y²z)] dx dy dz
= ∭_V(x² + y² + z²) dV = ∭_Va² dV
= a² ∭_VdV = a² (\(\frac {2}{3}\) πa³) = \(\frac {2}{3}\) πa⁵
∬_S[xz²dydx + (x²y – z³)dzdx + (2xy + y²z)dxdy = \(\frac {2}{3}\)πa⁵

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6. Which of the following is obtained by evaluating ∬_S curl \(\overrightarrow {F}.\hat{n}\) dS, where S is a closed surface and V is the Volume using Gauss Divergence Theorem?
a) 0
b) 1
c) -1
d) 2
View Answer

Answer: a
Explanation:
Gauss Divergence Theorem is
∬_S\(\overrightarrow {F}.\hat{n}\) dS = ∭_V∇.\(\overrightarrow {F}\)dV
where, V is the volume of the closed surface S.
Since, ∇.(curl \(\overrightarrow {F}\)) = 0, we get
∭_V∇.(curl \(\overrightarrow {F}\))dV = 0
∴ ∬_Scurl \(\overrightarrow {F}.\hat{n}\) dS = 0

7. Which of the following is obtained by evaluating ∬_S∅.\(\hat{n}\) dS, where S is a closed surface and V is the Volume using Gauss Divergence Theorem?
a) ∭_V∇∅ dV
b) ∭_V∇\(\hat{n}\)dV
c) ∭_V∇∅.∇∅ dV
d) 0
View Answer

Answer: a
Explanation:
Gauss Divergence Theorem is
∬_S\(\overrightarrow {F}.\hat{n}\) dS = ∭_V∇.\(\overrightarrow {F}\)dV
Put \(\overrightarrow {F}\) = ∅\(\overrightarrow {c}\), where \(\overrightarrow {c}\) is a constant vector
∬_S∅\(\overrightarrow {c}.\hat{n}\) dS = ∭_V∇.(∅\(\overrightarrow {c}\))dV
∬_S\(\overrightarrow {c}\).(∅\(\hat{n}\)) dS =∭_C\(\overrightarrow {c}\).∇∅ dV
Taking \(\overrightarrow {c}\) outside the integrals, we get
\(\overrightarrow {c}\).∬_S∅\(\hat{n}\) dS = \(\overrightarrow {c}\).∭_V∇∅ dV
∬_S∅\(\hat{n}\) dS = ∭_V∇∅ dV

8. Which of the following is obtained by evaluating ∬_S\(\overrightarrow {F}.\hat {n}\)dS ,where S is a closed surface and V is the Volume using Gauss Divergence Theorem if \(\overrightarrow {F}\) = x³\(\overrightarrow {i}\) + y³\(\overrightarrow {j}\) + z³ \(\overrightarrow {k}\) and the sphere
x²+ y² + z² = a²?
a) \(\frac {12}{5}\)πa⁵
b) \(\frac {1}{5}\)πa⁵
c) \(\frac {2}{5}\)πa⁵
d) \(\frac {12}{7}\)πa⁵
View Answer

Answer: a
Explanation:
div.\(\overrightarrow {F} = \frac {∂}{∂x}\)(x³) + \(\frac {∂}{∂y}\)(y³) + \(\frac {∂}{∂z}\)(z³) = 3(x² + y² + z²)
I = ∬_S\(\overrightarrow {F}.\overrightarrow{n}\)dS = ∭_Vdiv.\(\overrightarrow {F}\)dV
= ∭_V3(x²+ y² + z²)dx dy dz
x = rsin θ cos⁡∅; y = r sin⁡θ sin⁡∅; z = r cos⁡θ
x²+ y² + z² = r² and dxdydz = r² sin⁡θ dr dθ d∅
∴I = ∭_V3r².r² sin⁡θ dr dθ d∅
= 3\(\frac {a^5}{5}\).[-cos⁡θ]\(_0^π\) [∅]\(_0^2π\) = \(\frac {3}{5}\)a⁵[2][2π] = \(\frac {12}{5}\)πa⁵

9. Which of the following is obtained by evaluating ∬_S[x(y – z)\(\overrightarrow {i}\) + y(z – x)\(\overrightarrow {j}\) + z(x – y)\(\overrightarrow {k}\)]dS for any closed surface?
a) 0
b) e^r²\(\overrightarrow {r}\)
c) 2e^r⁴\(\overrightarrow {r}\)
d) 2\(\overrightarrow {r}\)
View Answer

Answer: a
Explanation:
Gauss Divergence Theorem is
∬_S\(\overrightarrow {F}.\hat{n}\) dS = ∭_V∇.\(\overrightarrow {F}\)dV
∇.\(\overrightarrow {F} = \bigg [ \overrightarrow {i} \frac {∂}{∂x} + \overrightarrow {j} \frac {∂}{∂y} + \overrightarrow {k} \frac {∂}{∂z} \bigg ]\).[x(y – z)\(\overrightarrow {i}\) + y(z – x)\(\overrightarrow {j}\) + z(x – y)\( \overrightarrow {k}\)]
= \(\frac {∂}{∂x}\)[x(y – z) ] + \(\frac {∂}{∂y}\)[y(z – x) ] + \(\frac {∂}{∂z}\)[z(x – y)]
= y – z + z – x + x – y = 0
∬_S[x(y – z) \(\overrightarrow {i}\)y(z – x)\(\overrightarrow {j}\) + z(x – y)\(\overrightarrow {k}\)]dS = ∭_V0 dV = 0

10. Which of the following is obtained by evaluating ∬_S\(\overrightarrow {r}.\overrightarrow {dS}\) where S is the surface of tetrahedron whose vertices are (0,0,0), (1,0,0), (0,1,0) and (0,0,1)?
a) 3
b) 2
c) 1
d) 0
View Answer

Answer: a
Explanation:
Gauss Divergence Theorem is
∬_S\(\overrightarrow {r}.\overrightarrow{dS}\) = ∭_V∇.\(\overrightarrow {r}\)dV
= ∭_V(\(\overrightarrow {i} \frac {∂}{∂x} + \overrightarrow {j} \frac {∂}{∂y} + \overrightarrow {k} \frac {∂}{∂z}\)).(x\(\overrightarrow {i}\) + y\(\overrightarrow {j}\) + z\(\overrightarrow {k} \))dV
= ∭_V(1 + 1 + 1) dV
= 3∭_VdV = 3\(\int {_0^1}\)\(\int {_0^1}\)\(\int {_0^1}\)dxdydz = 3(1)(1)(1) = 3

11. Which of the following is obtained by evaluating ∬_S\(\overrightarrow {r}.\hat{n}\) dS, where S is the surface of the sphere x² + y² + z² = 9?
a) 108π
b) 96π
c) 32π
d) 4π
View Answer

12. Which of the following is obtained by evaluating ∫_C(yz dx + zx dy + xy dz) where C is the curve x² + y² = 1, z = y²?
a) \(\overrightarrow {0}\)
b) e^r²\(\overrightarrow {r}\)
c) 2e^r⁴\(\overrightarrow {r}\)
d) 1
View Answer

Answer: a
Explanation:
Let (yz dx + zx dy + xy dz) = (yz\(\overrightarrow {i}\) + zx\(\overrightarrow {j}\) + xy\(\overrightarrow {k}\)).(\(\overrightarrow {i}\)dx + \(\overrightarrow {j}\)dy + \(\overrightarrow {k}\)dz)
= \(\overrightarrow {F}.\overrightarrow {dr}\)
when \(\overrightarrow {F}\) = (yz\(\overrightarrow {i}\) + zx\(\overrightarrow {j}\) + xy\(\overrightarrow {k}\))
∫_C(yz dx + zx dy + xy dz) = ∫_C\(\overrightarrow {F}.\overrightarrow {dr}\) = ∬_S(∇X\(\overrightarrow {F} ).\hat{n}\) dS
∇ X \(\overrightarrow {F} = \begin{vmatrix} \overrightarrow {i} & \overrightarrow {j} & \overrightarrow {k} \\
\frac {∂}{∂x} & \frac {∂}{∂y} & \frac {∂}{∂z} \\
yz & zx & xy \end{vmatrix}\)
= \(\overrightarrow {i}\)[x – x] – \(\overrightarrow {j}\)[y – y] + \(\overrightarrow {k}\)[z – z]
= \(\overrightarrow {0}\)

13. Which of the following is obtained by evaluating ∫_C\(\overrightarrow {r}.\overrightarrow {dr}\)?
a) 0
b) 1
c) 2
d) 3
View Answer

Answer: a
Explanation:
By Stokes theorem,
∫_C\(\overrightarrow {F} . \overrightarrow {dr}\) = ∬_S(∇ X \(\overrightarrow {F}).\hat{n}\) dS
Let \(\overrightarrow {F} = \overrightarrow {r}\) then ∇ X \(\overrightarrow {r} = \begin{vmatrix} \overrightarrow {i} & \overrightarrow {j} & \overrightarrow {k} \\
\frac {∂}{∂x} & \frac {∂}{∂y} & \frac {∂}{∂z} \\
yz & zx & xy \end{vmatrix}\) = 0
∴ ∫_C\(\overrightarrow {r}.\overrightarrow {dr}\) = 0

14. Which of the following is obtained by evaluating ∫_C(yz\(\overrightarrow {i}\) + xz\(\overrightarrow {j}\) + xy\(\overrightarrow {k}\)).\(\overrightarrow {dr}\) where C is the boundary of a surface S?
a) 0
b) 1
c) 2
d) 3
View Answer

Answer: a
Explanation:
Given: \(\overrightarrow {F}\) = yz\(\overrightarrow {i}\) + xz\(\overrightarrow {j}\) + xy\(\overrightarrow {k}\)
By Stoke’s theorem,
∫_C\(\overrightarrow {F} . \overrightarrow {dr}\) = ∬_S(∇ X \(\overrightarrow {F} ).\hat{n}\) dS
∇ X \(\overrightarrow {F} = \begin{vmatrix} \overrightarrow {i} & \overrightarrow {j} & \overrightarrow {k} \\
\frac {∂}{∂x} & \frac {∂}{∂y} & \frac {∂}{∂z} \\
yz & zx & xy \end{vmatrix} = \overrightarrow {i}\)[x – x] – \(\overrightarrow {j}\)[y – y] + \(\overrightarrow {k}\)[z – z]
= \(\overrightarrow {0}\)
∴∫_C\(\overrightarrow {F} . \overrightarrow {dr}\) = 0

15. Which of the following is obtained by evaluating ∫_C∅ ∇∅.\(\overrightarrow {dr}\)?
a) 0
b) 1
c) 2
d) 3
View Answer

Answer: a
Explanation:
Using Stoke’s theorem for ∅∇∅
∫_C∅∇∅ \(\overrightarrow {dr}\) = ∬_Scurl (∅∇∅).\(\hat{n}\)ds
= ∬_S[∅ curl ∇∅ + ∇∅ X ∇∅].\(\hat{n}\) dS
= 0 [∴curl ∇∅ = \(\overrightarrow {0}\) and ∇∅ X ∇∅ = \(\overrightarrow {0}\)]

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