BE Civil Engineering (IOE, TU) Engineering Mathematics III (IOE, SH 501) Question Paper 2079 Nepal

Green's theorem. If $C$ is a positively oriented, piecewise-smooth, simple closed curve in the plane bounding a region $R$, and $P(x,y)$, $Q(x,y)$ have continuous first partial derivatives on $R$, then

$$\oint_C P\,dx + Q\,dy = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA.$$

Identify P and Q.

$$P = 3x^2 - 8y^2, \qquad Q = 4y - 6xy.$$ $$\frac{\partial Q}{\partial x} = -6y, \qquad \frac{\partial P}{\partial y} = -16y.$$ $$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = -6y - (-16y) = 10y.$$

Region of integration. The curves $y = x^2$ and $y = x$ meet where $x^2 = x \Rightarrow x = 0,\ 1$. For $0 \le x \le 1$ the line lies above the parabola, so the region is

$$R = \{ (x,y) : 0 \le x \le 1,\ x^2 \le y \le x \}.$$

 y
 |        y = x
 |       /
 |      /  .· y = x^2
 |     / .·
 |    /.·
 |___/_____________ x
    0          1

Set up and evaluate the double integral.

$$\iint_R 10y\, dA = \int_0^1 \int_{x^2}^{x} 10y\, dy\, dx = \int_0^1 \left[ 5y^2 \right]_{x^2}^{x} dx = \int_0^1 5\left( x^2 - x^4 \right) dx.$$ $$= 5\left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_0^1 = 5\left( \frac{1}{3} - \frac{1}{5} \right) = 5 \cdot \frac{5 - 3}{15} = 5 \cdot \frac{2}{15} = \frac{10}{15} = \frac{2}{3}.$$

Final answer.

$$\boxed{\oint_C (3x^2 - 8y^2)\,dx + (4y - 6xy)\,dy = \frac{2}{3} \approx 0.667.}$$

Gauss divergence theorem. For a vector field $\mathbf{F}$ with continuous partial derivatives over a closed region $V$ bounded by the surface $S$ with outward unit normal $\mathbf{n}$,

$$\iiint_V (\nabla \cdot \mathbf{F})\, dV = \oiint_S \mathbf{F} \cdot \mathbf{n}\, dS.$$

LHS — Volume integral.

$$\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(2x) + \frac{\partial}{\partial y}(3y) + \frac{\partial}{\partial z}(z) = 2 + 3 + 1 = 6.$$ $$\iiint_V 6\, dV = 6 \times (\text{volume of unit cube}) = 6 \times 1 = 6.$$

RHS — Surface integral over the 6 faces. Outward normals and flux $\mathbf{F}\cdot\mathbf{n}$ on each face:

Each face has area $1$. Summing the fluxes:

$$\oiint_S \mathbf{F}\cdot\mathbf{n}\, dS = 2 + 0 + 3 + 0 + 1 + 0 = 6.$$

Conclusion. LHS $= 6 =$ RHS, so the divergence theorem is verified.

$$\boxed{\iiint_V (\nabla\cdot\mathbf{F})\,dV = \oiint_S \mathbf{F}\cdot\mathbf{n}\,dS = 6.}$$

Apply the Laplace transform. Let $Y(s) = \mathcal{L}\{y(t)\}$. Using

$$\mathcal{L}\{y''\} = s^2 Y - s\,y(0) - y'(0), \qquad \mathcal{L}\{y'\} = sY - y(0), \qquad \mathcal{L}\{e^{-2t}\} = \frac{1}{s+2}.$$

With $y(0)=0$, $y'(0)=1$:

$$\left( s^2 Y - 0 - 1 \right) + 4\left( sY - 0 \right) + 3Y = \frac{1}{s+2}.$$ $$\left( s^2 + 4s + 3 \right) Y - 1 = \frac{1}{s+2}.$$ $$(s+1)(s+3)\, Y = 1 + \frac{1}{s+2} = \frac{s+3}{s+2}.$$ $$Y = \frac{s+3}{(s+2)(s+1)(s+3)} = \frac{1}{(s+1)(s+2)}.$$

The factor $(s+3)$ cancels.

Partial fractions.

$$\frac{1}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2}.$$

$1 = A(s+2) + B(s+1)$. At $s=-1$: $1 = A(1) \Rightarrow A = 1$. At $s=-2$: $1 = B(-1) \Rightarrow B = -1$.

$$Y = \frac{1}{s+1} - \frac{1}{s+2}.$$

Inverse transform. Using $\mathcal{L}^{-1}\{1/(s+a)\} = e^{-at}$:

$$y(t) = e^{-t} - e^{-2t}.$$

Check initial conditions. $y(0) = 1 - 1 = 0$ ✓. $y'(t) = -e^{-t} + 2e^{-2t}$, so $y'(0) = -1 + 2 = 1$ ✓.

Final answer.

$$\boxed{y(t) = e^{-t} - e^{-2t}.}$$

Cauchy's residue theorem. If $f(z)$ is analytic inside and on a simple closed contour $C$ except at a finite number of isolated singular points $z_1, z_2, \dots, z_n$ interior to $C$, then

$$\oint_C f(z)\, dz = 2\pi i \sum_{k=1}^{n} \operatorname{Res}_{z = z_k} f(z).$$

Locate singularities. $z^2 - 2z = z(z-2) = 0 \Rightarrow z = 0$ and $z = 2$. Both are simple poles. Since $|0| = 0 < 3$ and $|2| = 2 < 3$, both poles lie inside $C: |z| = 3$.

$$f(z) = \frac{z^2 + 1}{z(z-2)}.$$

Residue at $z = 0$.

$$\operatorname{Res}_{z=0} f = \lim_{z\to 0} z \cdot \frac{z^2+1}{z(z-2)} = \frac{0+1}{0-2} = -\frac{1}{2}.$$

Residue at $z = 2$.

$$\operatorname{Res}_{z=2} f = \lim_{z\to 2} (z-2) \cdot \frac{z^2+1}{z(z-2)} = \frac{2^2+1}{2} = \frac{5}{2}.$$

Apply the theorem.

$$\sum \operatorname{Res} = -\frac{1}{2} + \frac{5}{2} = \frac{4}{2} = 2.$$ $$\oint_C f(z)\, dz = 2\pi i \times 2 = 4\pi i.$$

Final answer.

$$\boxed{\oint_C \frac{z^2+1}{z^2-2z}\, dz = 4\pi i.}$$

Boundary and initial conditions.

• $u(0,t) = 0$, $u(L,t) = u(2,t) = 0$ (fixed ends),
• $u_t(x,0) = 0$ (released from rest),
• $u(x,0) = 5\sin\!\left(\frac{\pi x}{2}\right) - 2\sin(2\pi x)$.

Separation of variables. Let $u(x,t) = X(x)\,T(t)$. Substituting gives $X T'' = c^2 X'' T$, hence

$$\frac{T''}{c^2 T} = \frac{X''}{X} = -\lambda^2.$$ $$X'' + \lambda^2 X = 0, \qquad T'' + c^2 \lambda^2 T = 0.$$

Spatial solution with fixed ends. $X(0) = X(2) = 0$ gives $X_n(x) = \sin\!\left(\frac{n\pi x}{L}\right) = \sin\!\left(\frac{n\pi x}{2}\right)$, with $\lambda_n = \frac{n\pi}{2}$, $n = 1, 2, 3, \dots$

Temporal solution (rest start). $T'' + c^2\lambda_n^2 T = 0$ gives $T_n(t) = a_n \cos(c\lambda_n t) + b_n \sin(c\lambda_n t)$. The rest condition $u_t(x,0)=0$ forces $b_n = 0$, so $T_n(t) = a_n \cos(c\lambda_n t)$.

General solution.

$$u(x,t) = \sum_{n=1}^{\infty} a_n \sin\!\left(\frac{n\pi x}{2}\right) \cos\!\left(\frac{3 n\pi}{2} t\right).$$

Match the initial displacement. At $t=0$:

$$\sum_{n=1}^\infty a_n \sin\!\left(\frac{n\pi x}{2}\right) = 5\sin\!\left(\frac{\pi x}{2}\right) - 2\sin(2\pi x).$$

Write $\sin(2\pi x) = \sin\!\left(\frac{4\pi x}{2}\right)$, which is the $n = 4$ mode. By comparing modes:

• $n = 1$: $a_1 = 5$,
• $n = 4$: $a_4 = -2$,
• all other $a_n = 0$.

Final solution.

$$\boxed{u(x,t) = 5\sin\!\left(\frac{\pi x}{2}\right)\cos\!\left(\frac{3\pi}{2}t\right) - 2\sin(2\pi x)\cos(6\pi t).}$$

(The frequencies follow from $c\lambda_n$: for $n=1$, $c\lambda_1 = 3\cdot\frac{\pi}{2} = \frac{3\pi}{2}$; for $n=4$, $c\lambda_4 = 3\cdot 2\pi = 6\pi$.)

Test for irrotationality: compute the curl.

$$\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial_x & \partial_y & \partial_z \\ 2xy+z^3 & x^2 & 3xz^2 \end{vmatrix}.$$

With $F_1 = 2xy+z^3$, $F_2 = x^2$, $F_3 = 3xz^2$:

• $\mathbf{i}$ component: $\partial_y F_3 - \partial_z F_2 = 0 - 0 = 0.$
• $\mathbf{j}$ component: $\partial_z F_1 - \partial_x F_3 = 3z^2 - 3z^2 = 0.$
• $\mathbf{k}$ component: $\partial_x F_2 - \partial_y F_1 = 2x - 2x = 0.$

$$\nabla \times \mathbf{F} = \mathbf{0}.$$

Hence $\mathbf{F}$ is irrotational (conservative).

Find the potential $\phi$. We need $\nabla\phi = \mathbf{F}$, i.e.

$$\frac{\partial\phi}{\partial x} = 2xy + z^3, \quad \frac{\partial\phi}{\partial y} = x^2, \quad \frac{\partial\phi}{\partial z} = 3xz^2.$$

Integrate the first w.r.t. $x$:

$$\phi = x^2 y + x z^3 + g(y,z).$$

Differentiate w.r.t. $y$: $\partial_y\phi = x^2 + g_y = x^2 \Rightarrow g_y = 0 \Rightarrow g = h(z).$

Differentiate w.r.t. $z$: $\partial_z\phi = 3xz^2 + h'(z) = 3xz^2 \Rightarrow h'(z) = 0 \Rightarrow h = C.$

Scalar potential.

$$\boxed{\phi(x,y,z) = x^2 y + x z^3 + C.}$$

Properties used.

1. Basic transform: $\mathcal{L}\{\cos(at)\} = \dfrac{s}{s^2 + a^2}$.
2. First shifting theorem: $\mathcal{L}\{e^{at} g(t)\} = G(s-a)$ where $G(s) = \mathcal{L}\{g(t)\}$.
3. Power rule: $\mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}}$.

Term 1: $e^{-3t}\cos(2t)$.

$$\mathcal{L}\{\cos 2t\} = \frac{s}{s^2 + 4}.$$

Apply the shift $s \to s - (-3) = s + 3$:

$$\mathcal{L}\{e^{-3t}\cos 2t\} = \frac{s+3}{(s+3)^2 + 4}.$$

Term 2: $t^2 e^{4t}$.

$$\mathcal{L}\{t^2\} = \frac{2!}{s^3} = \frac{2}{s^3}.$$

Apply the shift $s \to s - 4$:

$$\mathcal{L}\{t^2 e^{4t}\} = \frac{2}{(s-4)^3}.$$

Combine (linearity).

$$\boxed{\mathcal{L}\{f(t)\} = \frac{s+3}{(s+3)^2 + 4} + \frac{2}{(s-4)^3}.}$$

Valid for $s > 4$ (intersection of regions of convergence $s>-3$ and $s>4$).

Set up the integral.

$$F_c(\omega) = \sqrt{\frac{2}{\pi}} \int_0^\infty e^{-ax}\cos(\omega x)\, dx.$$

Evaluate the standard integral. Use

$$\int_0^\infty e^{-ax}\cos(\omega x)\, dx = \frac{a}{a^2 + \omega^2}, \qquad a > 0.$$

Derivation (integration by parts twice, or real part of $\int_0^\infty e^{-(a-i\omega)x}dx$):

$$\int_0^\infty e^{-(a-i\omega)x}\, dx = \frac{1}{a - i\omega} = \frac{a + i\omega}{a^2 + \omega^2}.$$

Taking the real part gives $\dfrac{a}{a^2+\omega^2}$.

Result.

$$\boxed{F_c(\omega) = \sqrt{\frac{2}{\pi}}\cdot\frac{a}{a^2 + \omega^2}.}$$

Check. At $\omega = 0$, $F_c(0) = \sqrt{\frac{2}{\pi}}\cdot\frac{a}{a^2} = \sqrt{\frac{2}{\pi}}\cdot\frac{1}{a}$, which equals $\sqrt{\frac{2}{\pi}}\int_0^\infty e^{-ax}dx = \sqrt{\frac{2}{\pi}}\cdot\frac{1}{a}$ ✓.

Check harmonicity (Laplace's equation $u_{xx} + u_{yy} = 0$).

$$u_x = 3x^2 - 3y^2, \qquad u_{xx} = 6x.$$ $$u_y = -6xy, \qquad u_{yy} = -6x.$$ $$u_{xx} + u_{yy} = 6x - 6x = 0.$$

So $u$ is harmonic.

Find $v$ using the Cauchy–Riemann equations $u_x = v_y$ and $u_y = -v_x$.

From $v_y = u_x = 3x^2 - 3y^2$, integrate w.r.t. $y$:

$$v = 3x^2 y - y^3 + g(x).$$

Differentiate w.r.t. $x$: $v_x = 6xy + g'(x)$. The CR equation $v_x = -u_y = 6xy$ gives

$$6xy + g'(x) = 6xy \Rightarrow g'(x) = 0 \Rightarrow g(x) = C.$$

Harmonic conjugate.

$$\boxed{v(x,y) = 3x^2 y - y^3 + C.}$$

Resulting analytic function. With $C=0$,

$$f(z) = (x^3 - 3xy^2) + i(3x^2 y - y^3) = z^3.$$

Describe the region. The region $R$ in the first quadrant is bounded below by $y = 0$ (the $x$-axis), above by $y = x$, on the right by $x = 2$. It is the triangle with vertices $(0,0)$, $(2,0)$, $(2,2)$.

 y
 |        (2,2)
 |       /|
 |      / |
 |     /  |
 |    /   |
 |___/____|____ x
   (0,0) (2,0)

Set up limits. For $0 \le x \le 2$, $y$ runs from $0$ to $x$:

$$\iint_R xy\, dA = \int_0^2 \int_0^x xy\, dy\, dx.$$

Inner integral (w.r.t. $y$).

$$\int_0^x xy\, dy = x\left[ \frac{y^2}{2} \right]_0^x = x\cdot\frac{x^2}{2} = \frac{x^3}{2}.$$

Outer integral (w.r.t. $x$).

$$\int_0^2 \frac{x^3}{2}\, dx = \frac{1}{2}\left[ \frac{x^4}{4} \right]_0^2 = \frac{1}{2}\cdot\frac{16}{4} = \frac{1}{2}\cdot 4 = 2.$$

Final answer.

$$\boxed{\iint_R xy\, dA = 2.}$$

Given relation.

$$z = (x^2 + a)(y^2 + b). \qquad (1)$$

There are two arbitrary constants $a$ and $b$, so we expect a first-order PDE obtained by differentiating w.r.t. $x$ and $y$.

Partial derivatives. Let $p = \dfrac{\partial z}{\partial x}$ and $q = \dfrac{\partial z}{\partial y}$.

$$p = \frac{\partial z}{\partial x} = 2x\,(y^2 + b). \qquad (2)$$ $$q = \frac{\partial z}{\partial y} = (x^2 + a)\,2y. \qquad (3)$$

Eliminate the constants. From (2): $\ y^2 + b = \dfrac{p}{2x}$. From (3): $\ x^2 + a = \dfrac{q}{2y}$.

Substitute both into (1):

$$z = (x^2 + a)(y^2 + b) = \frac{q}{2y}\cdot\frac{p}{2x} = \frac{pq}{4xy}.$$

Required PDE.

$$\boxed{4xy\,z = pq, \qquad \text{i.e.}\quad 4xyz = \frac{\partial z}{\partial x}\,\frac{\partial z}{\partial y}.}$$

Check. Multiplying (2) and (3): $pq = 4xy(x^2+a)(y^2+b) = 4xy\cdot z$, confirming $pq = 4xyz$ ✓.