# Vector Integral Calculus Questions and Answers – Stokes and Gauss Divergence Theorem

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

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

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 = ∭V2r2dV
= ∭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

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)

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

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
= a2VdV = a2 ($$\frac {2}{3}$$ πa3) = $$\frac {2}{3}$$ πa5
S[xz2dydx + (x2y – z3)dzdx + (2xy + y2z)dxdy = $$\frac {2}{3}$$πa5

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

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

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

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) er2$$\overrightarrow {r}$$
c) 2er4$$\overrightarrow {r}$$
d) 2$$\overrightarrow {r}$$

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

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

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
We know that ∬S$$\overrightarrow {r}.\hat{n}$$ dS = 3V
= 3$$\big [ \frac {4}{3}$$π33$$\big ]$$ = 108π
[∴ V is the volume of the sphere]

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) er2$$\overrightarrow {r}$$
c) 2er4$$\overrightarrow {r}$$
d) 1

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

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

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

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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