PYQ on Vector Calculus | Engineering Mathematics

In examinations, questions from Vector Calculus are often asked in the form of evaluating line integrals, surface integrals, and volume integrals of vector fields. Unlike standard multivariable calculus, vector calculus focuses on vector-valued functions and their interaction with curves, surfaces, and volumes.

Short Question on Vector Calculus

Question 1: Let S be sphere x² + y² + z² = 1 then what is value of integral \int\int(xsiny , cos² x, 2z - z sin y).(x,y,z).

Question 2: Evaluate \int_s(4x \hat{i} - 2y^{2}\hat{j} + z^{2}\hat{k}).\hat{n}where S is bounded by x² + y² = 4, z = 0, and z = 3.

Question 3: Show that the area of an ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 is πzb using Green's Theorem.

Question 4: Evaluate \oint_{c} yzdx + zxdy + xydz where 'c' is the curve, x² + y² = 1, z = y².

Question 5: The directional derivative of f(x,y) = xy/√2(x+y) at (1,1) in the direction of the unit vector at an angle of \frac{\pi}{4} with the y-axis.

Question 6: Find the magnitude of the gradient for the function f(x, y, z) = x² + 3y² + z³ at the point (1, 1, 1).

Question 7: Evaluate the line integral \int_C (y^2 \, dx + x^2 \, dy) where C is the boundary of the region enclosed by the parabola y = x² and the line y = 1.

Question 8: Evaluate \iint_S (x^3 + y^3 + z^3) \, dS where S$is the plane x + y + z = 1 in the first octant.

Question 9: Use Divergence Theorem to evaluate \iint_S \mathbf{F} \cdot \mathbf{n} \, dS for \mathbf{F} = (x^2, y^2, z^2) where S is the surface of the cube bounded by x, y, z = 0 to 1.

Question 10: Find the gradient of f(x,y) = \ln(x^2 + y^2)at the point (1,2) and its magnitude.

Long Questions on Vector Calculus

Question 11: Evaluate the surface integral \iint_S (y z \, \hat{i} + x z \, \hat{j} + x y \, \hat{k}) \cdot \hat{n} \, dSwhere S is the surface of the cube bounded by x, y, and z = 0 to 1.

Question 12: Use Green's theorem to find the area enclosed by the ellipse \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

Check if you were right - full answer with solution below.

Short Question on Vector Calculus: Answers

Solution 1:

\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(x\sin y) + \frac{\partial}{\partial y}(\cos^2x) + \frac{\partial}{\partial z}(2z - z\sin y)= \sin y + 0 + (2 - \sin y) = 2
Therefore , \iiint_V (\nabla \cdot \vec{F}) , dV = 2

Volume of Sphere = 2 \times \frac{4}{3}\pi = \frac{8}{3}\pi

Solution 2:

\nabla \cdot \vec{F} = 4 - 4y + 2z

\iiint_V (\nabla \cdot \vec{F})\, dV = \int_0^3 \int_0^{2\pi} \int_0^2 (4 + 2z)r\, dr\, d\theta\, dz

= 2\pi \int_0^3 \left[\frac{r^2}{2}(4+2z)\right]_0^2 dz= 2\pi \int_0^3 (8 + 4z)\, dz = 2\pi [8z + 2z^2]_0^3 = 2\pi(24 + 18) = 84\pi

Solution 3:

By Green's theorem:

A = \oint_C x\,dy = -\oint_C y\,dx = \frac{1}{2}\oint_C (x\,dy - y\,dx)

x = a\cos t, \ y = b\sin t \Rightarrow A = \frac{1}{2}\int_0^{2\pi}(a\cos t)(b\cos t) - (b\sin t)(-a\sin t)\,dt = \pi ab

x = acos t, y = bsin t

A = \frac{1}{2}\int_0^{2\pi}(a\cos t)(b\cos t) - (b\sin t)(-a\sin t)\,dt = \pi ab

Solution 4:

dx = -sin tdt, dy = cos tdt

I = \int_0^{\pi/2} (\cos t \cdot \cos t - \sin t \cdot (-\sin t))dt

= \int_0^{\pi/2} (\cos^2t + \sin^2t)dt

= \frac{\pi}{2}

Solution 5:

x = cos t, y = sin t

dx = -sin tdt,

dy = cos tdt

I = \int_0^{2\pi}(\sin^2t(-\sin t) + \cos^2t(\cos t))dt

\int_0^{2\pi}(\cos^3t - \sin^3t)\,dt = 0

Solution 6:

\nabla \cdot \vec{F} = \frac{\partial(yz)}{\partial x} + \frac{\partial(zx)}{\partial y} + \frac{\partial(xy)}{\partial z}

0 + 0 + 0

= 0

\nabla \cdot \vec{F} = 0

Solution 7:

\nabla \times \vec{F} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\2xy & 3yz & 4zx\end{vmatrix}

(4z - 3z)\hat{i} - (4x - 2x)\hat{j} + (3y - 2y)\hat{k}

= (z, -2x ,y)

Solution 8:

\nabla \cdot \vec{F} = \frac{1}{r}\frac{\partial}{\partial r}(r F_r) + \frac{1}{r}\frac{\partial F_\theta}{\partial \theta}

\frac{1}{r}\frac{\partial}{\partial r}(r^4\sin\theta) + \frac{1}{r}\frac{\partial}{\partial \theta}(r^2\cos\theta)

= 4r²sinθ- r²/r(sin θ) = 3r²sin θ

Solution 9:

\nabla \times \vec{A} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\x^2 & y^2 & z^2\end{vmatrix}

( - 0)\hat{i} - (0 - 0 )\hat{j} + (0 - 0)\hat{k}

\nabla \times \vec{A} = 0

Solution 10:

\nabla \cdot \vec{F} = \iiint_V (\nabla \cdot \vec{F})\,dV

= 3 x Volume of cube

= 3 x 8

= 24

Hence flux through cube = 24

Long Question on Vector Calculus: Answers

Solution 11:

By the divergence Theorem,

\iint_{S}\mathbf{F}\cdot\mathbf{n}\,dS \;=\; \iiint_{V} \nabla\cdot\mathbf{F}\,dV

\nabla\cdot\mathbf{F}=\frac{\partial}{\partial x}(x^3)+\frac{\partial}{\partial y}(y^3)+\frac{\partial}{\partial z}(z^3)

3x² + 3y² + 3z² = 3(x² + y² + z²)

\iiint_V 3r^2\,dV = 3\int_{0}^{1}\int_{0}^{\pi}\int_{0}^{2\pi} r^2\,(r^2\sin\theta)\,d\phi\,d\theta\,dr

3\cdot 2\pi \int_{0}^{\pi}\sin\theta\,d\theta \int_{0}^{1} r^4\,dr

= =3\cdot 2\pi \cdot 2 \cdot \left[\frac{r^5}{5}\right]_{0}^{1}

=12\pi\cdot\frac{1}{5}=\frac{12\pi}{5}.

Therefore, \iint_{S}\mathbf{F}\cdot\mathbf{n}\,dS=\dfrac{12\pi}{5}\;

Solution 12:

Compute Curl:

\nabla\times\mathbf{F}=\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\[4pt]\partial_x & \partial_y & \partial_z\\[4pt]-y & x & z\end{vmatrix}

=(\partial_y z-\partial_z x)\mathbf{i}-(\partial_x z-\partial_z(-y))\mathbf{j}+(\partial_x x-\partial_y(-y))\mathbf{k}

Simplify:

\nabla\times\mathbf{F}=(0-0)i-(0-0)j+(1-(-1))k

=(0,0,2)

(\nabla\times\mathbf{F})\cdot\mathbf{n} = 2

So surface integral over the unit disk is

\iint_{S}2\,dS = 2 . (area of unit disk)

= 2\cdot\pi\cdot 1^2=2\pi