BE Civil Engineering (IOE, TU) Engineering Mathematics III (IOE, SH 501) Question Paper 2078 Nepal

Green's theorem states:

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

Here $P = 3x^2 - 8y^2$ and $Q = 4y - 6xy$.

Step 1 — Compute the partial derivatives.

$$\frac{\partial P}{\partial y} = -16y, \qquad \frac{\partial Q}{\partial x} = -6y$$ $$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = -6y - (-16y) = 10y$$

Step 2 — Region of integration. The curves $y = x^2$ and $y = x$ intersect where $x^2 = x \Rightarrow x = 0, 1$. For $0 \le x \le 1$, the line $y = x$ lies above the parabola $y = x^2$. So $x^2 \le y \le x$.

Step 3 — Evaluate the double integral.

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

Step 4 — Evaluate the line integral directly. $C = C_1 \cup C_2$.

Along $C_1$ ($y = x^2$, from $(0,0)$ to $(1,1)$): $dy = 2x\,dx$, $x: 0\to1$.

$$\int_{C_1} = \int_0^1 \left[(3x^2 - 8x^4) + (4x^2 - 6x\cdot x^2)(2x)\right]dx$$ $$= \int_0^1 \left[3x^2 - 8x^4 + 8x^3 - 12x^4\right]dx = \int_0^1 (3x^2 + 8x^3 - 20x^4)\,dx$$ $$= \left[x^3 + 2x^4 - 4x^5\right]_0^1 = 1 + 2 - 4 = -1$$

Along $C_2$ ($y = x$, from $(1,1)$ to $(0,0)$): $dy = dx$, $x: 1\to0$.

$$\int_{C_2} = \int_1^0 \left[(3x^2 - 8x^2) + (4x - 6x^2)\right]dx = \int_1^0 (-5x^2 + 4x - 6x^2)\,dx$$ $$= \int_1^0 (-11x^2 + 4x)\,dx = \left[-\frac{11x^3}{3} + 2x^2\right]_1^0 = 0 - \left(-\frac{11}{3} + 2\right) = \frac{11}{3} - 2 = \frac{5}{3}$$

Step 5 — Total line integral.

$$\oint_C = -1 + \frac{5}{3} = \frac{2}{3}$$

Since the line integral $= \frac{2}{3}$ equals the double integral $= \frac{2}{3}$, Green's theorem is verified. $\boxed{\oint_C = \dfrac{2}{3}}$

Divergence theorem. For a vector field $\vec{F}$ with continuous first partial derivatives over a region $V$ bounded by a closed surface $S$ with outward unit normal $\hat{n}$:

$$\iint_S \vec{F}\cdot\hat{n}\,dS = \iiint_V (\nabla\cdot\vec{F})\,dV$$

Step 1 — Divergence.

$$\nabla\cdot\vec{F} = \frac{\partial}{\partial x}(4x) + \frac{\partial}{\partial y}(-2y^2) + \frac{\partial}{\partial z}(z^2) = 4 - 4y + 2z$$

Step 2 — Set up the volume integral over the cylinder $x^2+y^2\le4$, $0\le z\le3$. Use cylindrical coordinates: $x=r\cos\theta$, $y=r\sin\theta$, $dV = r\,dr\,d\theta\,dz$, with $0\le r\le2$, $0\le\theta\le2\pi$, $0\le z\le3$.

$$\iiint_V (4 - 4y + 2z)\,dV = \iiint_V (4 - 4r\sin\theta + 2z)\,r\,dr\,d\theta\,dz$$

Step 3 — The $-4r\sin\theta$ term vanishes because $\int_0^{2\pi}\sin\theta\,d\theta = 0$. So:

$$= \int_0^3\!\int_0^{2\pi}\!\int_0^2 (4 + 2z)\,r\,dr\,d\theta\,dz$$

Step 4 — Integrate. $\int_0^2 r\,dr = 2$, $\int_0^{2\pi} d\theta = 2\pi$.

$$= 2\cdot 2\pi \int_0^3 (4 + 2z)\,dz = 4\pi\left[4z + z^2\right]_0^3 = 4\pi(12 + 9) = 4\pi\cdot 21 = 84\pi$$

Final answer: $\boxed{\displaystyle\iint_S \vec{F}\cdot\hat{n}\,dS = 84\pi \approx 263.89}$

Step 1 — Transform both sides. Let $Y(s) = \mathcal{L}\{y\}$. Using $\mathcal{L}\{y''\} = s^2Y - sy(0) - y'(0)$ and $\mathcal{L}\{y'\} = sY - y(0)$:

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

Step 2 — Solve for $Y$. Note $s^2+4s+3 = (s+1)(s+3)$.

$$(s+1)(s+3)\,Y = s + 4 + \frac{1}{s+2}$$ $$Y = \frac{s+4}{(s+1)(s+3)} + \frac{1}{(s+1)(s+2)(s+3)}$$

Step 3 — Partial fractions, first term.

$$\frac{s+4}{(s+1)(s+3)} = \frac{A}{s+1} + \frac{B}{s+3}$$

At $s=-1$: $A = \frac{-1+4}{-1+3} = \frac{3}{2}$. At $s=-3$: $B = \frac{-3+4}{-3+1} = \frac{1}{-2} = -\frac{1}{2}$.

Step 4 — Partial fractions, second term.

$$\frac{1}{(s+1)(s+2)(s+3)} = \frac{C}{s+1} + \frac{D}{s+2} + \frac{E}{s+3}$$

At $s=-1$: $C = \frac{1}{(1)(2)} = \frac{1}{2}$. At $s=-2$: $D = \frac{1}{(-1)(1)} = -1$. At $s=-3$: $E = \frac{1}{(-2)(-1)} = \frac{1}{2}$.

Step 5 — Combine coefficients.

$$Y = \frac{\frac{3}{2}+\frac{1}{2}}{s+1} + \frac{-1}{s+2} + \frac{-\frac{1}{2}+\frac{1}{2}}{s+3} = \frac{2}{s+1} - \frac{1}{s+2} + \frac{0}{s+3}$$ $$Y = \frac{2}{s+1} - \frac{1}{s+2}$$

Step 6 — Inverse transform. Using $\mathcal{L}^{-1}\{1/(s+a)\} = e^{-at}$:

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

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

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 $z_1, z_2, \dots, z_n$ inside $C$, then

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

Step 1 — Locate singularities. $f(z) = \dfrac{z^2+1}{(z-1)(z-3)}$ has simple poles at $z = 1$ and $z = 3$.

Step 2 — Which lie inside $C: |z|=2$? $|1| = 1 < 2$ (inside); $|3| = 3 > 2$ (outside). Only $z = 1$ contributes.

Step 3 — Residue at the simple pole $z = 1$.

$$\operatorname{Res}_{z=1} f(z) = \lim_{z\to1} (z-1)\frac{z^2+1}{(z-1)(z-3)} = \frac{z^2+1}{z-3}\bigg|_{z=1} = \frac{1+1}{1-3} = \frac{2}{-2} = -1$$

Step 4 — Apply the theorem.

$$\oint_C f(z)\,dz = 2\pi i \cdot (-1) = -2\pi i$$

Final answer: $\boxed{\displaystyle\oint_C \frac{z^2+1}{(z-1)(z-3)}\,dz = -2\pi i}$

Boundary/initial conditions.

• $u(0,t) = 0$, $u(\pi,t) = 0$ (fixed ends)
• $u(x,0) = 3\sin x - 2\sin 4x$ (initial shape)
• $u_t(x,0) = 0$ (released from rest)

Step 1 — Separation of variables. Let $u(x,t) = X(x)T(t)$. Substituting into $u_{tt} = c^2 u_{xx}$:

$$X T'' = c^2 X'' T \;\Rightarrow\; \frac{T''}{c^2 T} = \frac{X''}{X} = -\lambda^2$$

This gives $X'' + \lambda^2 X = 0$ and $T'' + c^2\lambda^2 T = 0$.

Step 2 — Solve the spatial problem. $X(x) = A\cos\lambda x + B\sin\lambda x$. The fixed ends require $X(0)=0 \Rightarrow A=0$, and $X(\pi)=0 \Rightarrow B\sin\lambda\pi = 0$. Non-trivial solutions need $\lambda\pi = n\pi$, i.e. $\lambda = n$ for $n = 1,2,3,\dots$. So $X_n(x) = \sin nx$.

Step 3 — Solve the time problem. $T_n'' + c^2 n^2 T_n = 0$ gives $T_n(t) = C_n\cos(nct) + D_n\sin(nct)$.

Step 4 — Apply zero initial velocity. $u_t(x,0)=0$ forces $D_n = 0$. The general solution:

$$u(x,t) = \sum_{n=1}^{\infty} C_n \sin(nx)\cos(nct)$$

Step 5 — Match initial displacement. At $t=0$:

$$u(x,0) = \sum_{n=1}^{\infty} C_n \sin(nx) = 3\sin x - 2\sin 4x$$

Matching coefficients (the series is already in eigenfunction form): $C_1 = 3$, $C_4 = -2$, all other $C_n = 0$.

Final solution:

$$\boxed{u(x,t) = 3\sin x\,\cos(ct) - 2\sin 4x\,\cos(4ct)}$$

This represents the superposition of the first and fourth normal modes of vibration, each oscillating at its own frequency $\omega_n = nc$.

Step 1 — Identify the region. The inner limits give $\sqrt{x} \le y \le 2$ and the outer give $0 \le x \le 4$. Equivalently, the region is bounded by $y = \sqrt{x}$ (i.e. $x = y^2$), $y = 2$, and $x = 0$.

 y
 2 +--------+   (4,2)
   |       /
   |     /  y = sqrt(x)  =>  x = y^2
   |   /
   | /
 0 +--------+ x
   0        4

Step 2 — Reverse the order. For a fixed $y$ between $0$ and $2$, $x$ runs from $0$ to $y^2$ (since $y = \sqrt{x} \Rightarrow x = y^2$ is the right boundary, and $x=0$ the left). Thus:

$$\int_0^2 \int_0^{y^2} \frac{x}{y^5+1}\,dx\,dy$$

Step 3 — Integrate with respect to $x$.

$$\int_0^{y^2} \frac{x}{y^5+1}\,dx = \frac{1}{y^5+1}\cdot\frac{x^2}{2}\bigg|_0^{y^2} = \frac{1}{y^5+1}\cdot\frac{y^4}{2} = \frac{y^4}{2(y^5+1)}$$

Step 4 — Integrate with respect to $y$. Let $u = y^5 + 1$, $du = 5y^4\,dy$.

$$\int_0^2 \frac{y^4}{2(y^5+1)}\,dy = \frac{1}{2}\cdot\frac{1}{5}\int \frac{du}{u} = \frac{1}{10}\ln(y^5+1)\bigg|_0^2$$ $$= \frac{1}{10}\left[\ln(33) - \ln(1)\right] = \frac{1}{10}\ln 33$$

Final answer: $\boxed{\dfrac{\ln 33}{10} \approx 0.3497}$

(Changing the order was essential since $\int 1/(y^5+1)\,dy$ in the original order has no elementary form.)

Part (a) — Gradient.

$$\nabla\phi = \frac{\partial\phi}{\partial x}\hat{i} + \frac{\partial\phi}{\partial y}\hat{j} + \frac{\partial\phi}{\partial z}\hat{k}$$ $$\frac{\partial\phi}{\partial x} = 2xy + z^2,\quad \frac{\partial\phi}{\partial y} = x^2 + 2yz,\quad \frac{\partial\phi}{\partial z} = y^2 + 2zx$$

At $P(1,2,-1)$:

$$\frac{\partial\phi}{\partial x} = 2(1)(2) + (-1)^2 = 4 + 1 = 5$$ $$\frac{\partial\phi}{\partial y} = (1)^2 + 2(2)(-1) = 1 - 4 = -3$$ $$\frac{\partial\phi}{\partial z} = (2)^2 + 2(-1)(1) = 4 - 2 = 2$$ $$\boxed{\nabla\phi\big|_P = 5\hat{i} - 3\hat{j} + 2\hat{k}}$$

Part (b) — Directional derivative. The unit vector along $\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}$:

$$|\vec{a}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3$$ $$\hat{a} = \frac{1}{3}(2\hat{i} - \hat{j} + 2\hat{k})$$

Directional derivative:

$$D_{\hat{a}}\phi = \nabla\phi\cdot\hat{a} = (5\hat{i} - 3\hat{j} + 2\hat{k})\cdot\frac{1}{3}(2\hat{i} - \hat{j} + 2\hat{k})$$ $$= \frac{1}{3}\left[(5)(2) + (-3)(-1) + (2)(2)\right] = \frac{1}{3}(10 + 3 + 4) = \frac{17}{3}$$

Final answer: $\boxed{D_{\hat{a}}\phi = \dfrac{17}{3} \approx 5.67}$

Part (a). Start from $\mathcal{L}\{\sin 2t\} = \dfrac{2}{s^2+4}$.

Apply the first shifting theorem $\mathcal{L}\{e^{-3t}g(t)\} = G(s+3)$:

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

Apply multiplication by $t$: $\mathcal{L}\{t\,h(t)\} = -\dfrac{d}{ds}H(s)$. Let $H(s) = \dfrac{2}{(s+3)^2+4}$.

$$-\frac{d}{ds}\left[2\big((s+3)^2+4\big)^{-1}\right] = -2\cdot\big(-1\big)\big((s+3)^2+4\big)^{-2}\cdot 2(s+3)$$ $$= \frac{4(s+3)}{\big[(s+3)^2+4\big]^2}$$ $$\boxed{\mathcal{L}\{t\,e^{-3t}\sin 2t\} = \frac{4(s+3)}{\big[(s+3)^2 + 4\big]^2}}$$

Part (b) — Convolution. Write $\dfrac{1}{s(s^2+4)} = \dfrac{1}{s}\cdot\dfrac{1}{s^2+4}$.

$$\mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1, \qquad \mathcal{L}^{-1}\left\{\frac{1}{s^2+4}\right\} = \frac{1}{2}\sin 2t$$

By the convolution theorem, $\mathcal{L}^{-1}\{F(s)G(s)\} = (f*g)(t) = \int_0^t f(\tau)g(t-\tau)\,d\tau$:

$$\mathcal{L}^{-1}\left\{\frac{1}{s(s^2+4)}\right\} = \int_0^t 1\cdot\frac{1}{2}\sin 2(t-\tau)\,d\tau$$ $$= \frac{1}{2}\left[\frac{\cos 2(t-\tau)}{2}\right]_0^t = \frac{1}{4}\left[\cos 0 - \cos 2t\right] = \frac{1}{4}(1 - \cos 2t)$$

Final answer: $\boxed{\mathcal{L}^{-1}\left\{\dfrac{1}{s(s^2+4)}\right\} = \dfrac{1}{4}(1 - \cos 2t)}$

Step 1 — Split the integral at $x=0$ since $|x|$ changes form. For $x\ge0$, $|x|=x$; for $x<0$, $|x|=-x$.

$$F(\omega) = \int_{-\infty}^{0} e^{ax}e^{-i\omega x}\,dx + \int_{0}^{\infty} e^{-ax}e^{-i\omega x}\,dx$$

Step 2 — Evaluate each piece.

Second integral: $\displaystyle\int_0^\infty e^{-(a+i\omega)x}\,dx = \left[\frac{e^{-(a+i\omega)x}}{-(a+i\omega)}\right]_0^\infty = \frac{1}{a+i\omega}$ (since $a>0$ the upper limit vanishes).

First integral: $\displaystyle\int_{-\infty}^0 e^{(a-i\omega)x}\,dx = \left[\frac{e^{(a-i\omega)x}}{a-i\omega}\right]_{-\infty}^0 = \frac{1}{a-i\omega}$.

Step 3 — Combine.

$$F(\omega) = \frac{1}{a-i\omega} + \frac{1}{a+i\omega} = \frac{(a+i\omega) + (a-i\omega)}{(a-i\omega)(a+i\omega)} = \frac{2a}{a^2+\omega^2}$$ $$\boxed{F(\omega) = \frac{2a}{a^2 + \omega^2}}$$

Step 4 — Evaluate the integral. By the inverse Fourier transform, $f(0) = \dfrac{1}{2\pi}\int_{-\infty}^\infty F(\omega)\,d\omega$. Since $f(0) = e^0 = 1$:

$$1 = \frac{1}{2\pi}\int_{-\infty}^\infty \frac{2a}{a^2+\omega^2}\,d\omega \;\Rightarrow\; \int_{-\infty}^\infty \frac{1}{a^2+\omega^2}\,d\omega = \frac{\pi}{a}$$

For $a = 2$:

$$\boxed{\int_{-\infty}^\infty \frac{1}{4+\omega^2}\,d\omega = \frac{\pi}{2} \approx 1.5708}$$

Step 1 — Check harmonicity. A function is harmonic if $u_{xx} + u_{yy} = 0$.

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

So $u$ is harmonic.

Step 2 — Use the Cauchy–Riemann equations to find $v$: $v_y = u_x$ and $v_x = -u_y$.

From $v_y = u_x = 3x^2 - 3y^2$, integrate with respect to $y$:

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

Step 3 — Differentiate and match. $v_x = 6xy + g'(x)$. But CR requires $v_x = -u_y = -(-6xy+2) = 6xy - 2$. So:

$$6xy + g'(x) = 6xy - 2 \;\Rightarrow\; g'(x) = -2 \;\Rightarrow\; g(x) = -2x + C$$

Taking $C = 0$:

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

Step 4 — Form $f(z)$.

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

Group the cubic terms: $x^3 - 3xy^2 + i(3x^2 y - y^3) = (x+iy)^3 = z^3$. Group the linear terms: $2y - 2ix = -2i(x + iy) = -2iz$ (since $-2i(x+iy) = -2ix - 2i^2y = -2ix + 2y$).

$$\boxed{f(z) = z^3 - 2iz}$$

Check: $f(z) = z^3 - 2iz$ is a polynomial, hence entire (analytic everywhere), confirming the result.

Part (a) — Classification. For a PDE of the form $A u_{xx} + B u_{xy} + C u_{yy} + \dots = 0$, classify using the discriminant $B^2 - 4AC$:

• $B^2 - 4AC > 0$: hyperbolic
• $B^2 - 4AC = 0$: parabolic
• $B^2 - 4AC < 0$: elliptic

Here $A = 4$, $B = 4$, $C = 1$:

$$B^2 - 4AC = 4^2 - 4(4)(1) = 16 - 16 = 0$$

Since the discriminant is zero, the PDE is parabolic.

Part (b) — Eliminate $a$ and $b$. Given $z = (x^2+a)(y^2+b)$.

Partial derivatives:

$$p = \frac{\partial z}{\partial x} = 2x(y^2 + b) \;\Rightarrow\; y^2 + b = \frac{p}{2x}$$ $$q = \frac{\partial z}{\partial y} = 2y(x^2 + a) \;\Rightarrow\; x^2 + a = \frac{q}{2y}$$

Multiply the two original factors:

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

Therefore:

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