BE Civil Engineering (IOE, TU) Engineering Mathematics III (IOE, SH 501) Question Paper 2080 Nepal

Green's theorem. If $C$ is a positively oriented, piecewise-smooth, simple closed curve enclosing a region $R$ in the plane, 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 $Q$.

$$P = 3x^2 - 8y^2,\qquad Q = 4y - 6xy.$$

Compute partials.

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

Region of integration. The curves $y=\sqrt{x}$ and $y=x$ meet where $\sqrt{x}=x \Rightarrow x = x^2 \Rightarrow x(x-1)=0$, i.e. at $x=0$ and $x=1$. On $0\le x\le 1$, $\sqrt{x}\ge x$, so the region is

$$R: \quad 0 \le x \le 1,\qquad x \le y \le \sqrt{x}.$$

 y
 1 +        * (1,1)
   |     .' /
   |   .'  /   y = sqrt(x) (upper)
   | .'   /    y = x       (lower)
 0 +-----+----- x
   0     1

Set up the double integral.

$$\iint_R 10y\,dA = \int_0^1 \int_x^{\sqrt{x}} 10y\,dy\,dx.$$

Inner integral:

$$\int_x^{\sqrt{x}} 10y\,dy = 10\cdot\frac{y^2}{2}\Big|_x^{\sqrt{x}} = 5\left[(\sqrt{x})^2 - x^2\right] = 5\left(x - x^2\right).$$

Outer integral:

$$\int_0^1 5\left(x - x^2\right)\,dx = 5\left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = 5\left(\frac{1}{2} - \frac{1}{3}\right) = 5\cdot\frac{1}{6} = \frac{5}{6}.$$

Result.

$$\boxed{\oint_C P\,dx + Q\,dy = \dfrac{5}{6} \approx 0.8333}$$

Divergence theorem. For a vector field $\mathbf{F}$ with continuous partial derivatives on a solid region $V$ bounded by the closed surface $S$ (outward normal $\mathbf{n}$),

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

Compute the divergence.

$$\nabla\cdot\mathbf{F} = \frac{\partial}{\partial x}(2x) + \frac{\partial}{\partial y}(3y) + \frac{\partial}{\partial z}(z^2) = 2 + 3 + 2z = 5 + 2z.$$

Set up the volume integral over the cube $0\le x\le 2,\ 0\le y\le 2,\ 0\le z\le 2$:

$$\iiint_V (5 + 2z)\,dV = \int_0^2\!\int_0^2\!\int_0^2 (5 + 2z)\,dx\,dy\,dz.$$

The integrand does not depend on $x$ or $y$, so integrating over $x$ and $y$ gives a factor $2\times 2 = 4$:

$$= 4\int_0^2 (5 + 2z)\,dz.$$

Evaluate the $z$-integral:

$$\int_0^2 (5 + 2z)\,dz = \left[5z + z^2\right]_0^2 = (10 + 4) - 0 = 14.$$

Therefore

$$\iint_S \mathbf{F}\cdot \mathbf{n}\,dS = 4 \times 14 = 56.$$

Check (direct, face-by-face). Flux through each pair of opposite faces:

• $x$-faces: at $x=2$, flux $=\int\!\int 2(2)\,dy\,dz = 4\cdot4=16$; at $x=0$, $2(0)=0$ outward $(-x)$ contributes $0$. Net $=16$.
• $y$-faces: at $y=2$, $3(2)=6$, flux $=6\cdot4=24$; at $y=0$, $0$. Net $=24$.
• $z$-faces: at $z=2$, $z^2=4$, flux $=4\cdot4=16$; at $z=0$, $0$. Net $=16$.

Total $=16+24+16 = 56$, confirming the answer.

Result.

$$\boxed{\iint_S \mathbf{F}\cdot \mathbf{n}\,dS = 56}$$

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

$$\mathcal{L}\{y''\} = s^2 Y - s\,y(0) - y'(0),\qquad \mathcal{L}\{y'\} = sY - y(0),$$

and $\mathcal{L}\{e^{-2t}\} = \dfrac{1}{s+2}$:

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

Collect terms:

$$(s^2 + 4s + 3)Y - s - 4 = \frac{1}{s+2}.$$ $$(s^2 + 4s + 3)Y = s + 4 + \frac{1}{s+2} = \frac{(s+4)(s+2) + 1}{s+2} = \frac{s^2 + 6s + 9}{s+2} = \frac{(s+3)^2}{s+2}.$$

Since $s^2 + 4s + 3 = (s+1)(s+3)$,

$$Y = \frac{(s+3)^2}{(s+2)(s+1)(s+3)} = \frac{s+3}{(s+1)(s+2)}.$$

Partial fractions.

$$\frac{s+3}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2}.$$

Multiply out: $s+3 = A(s+2) + B(s+1)$.

• $s=-1$: $-1+3 = A(1) \Rightarrow A = 2.$
• $s=-2$: $-2+3 = B(-1) \Rightarrow B = -1.$

So

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

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

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

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

Result.

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

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

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

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

The contour is $|z|=2$ (radius 2 about the origin):

• $z=1$: $|1| = 1 < 2$ — inside $C$.
• $z=3$: $|3| = 3 > 2$ — outside $C$.

Only $z=1$ contributes.

Residue at the simple pole $z=1$.

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

Apply the theorem.

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

Result.

$$\boxed{\oint_{|z|=2} \frac{z^2+1}{(z-1)(z-3)}\,dz = -2\pi i}$$

Separation of variables. Seek $u(x,t) = X(x)\,T(t)$. Substituting into the wave equation:

$$X T'' = c^2 X'' T \;\Rightarrow\; \frac{T''}{c^2 T} = \frac{X''}{X} = -\lambda \;(\text{const}).$$

This gives two ODEs:

$$X'' + \lambda X = 0,\qquad T'' + \lambda c^2 T = 0.$$

Spatial problem and boundary conditions. $u(0,t)=0 \Rightarrow X(0)=0$; $u(\pi,t)=0 \Rightarrow X(\pi)=0$. For nontrivial solutions take $\lambda = n^2$ ($n=1,2,\dots$):

$$X_n(x) = \sin(nx),\qquad n=1,2,3,\dots$$

(the eigenvalues are $\lambda_n = n^2$ because $L=\pi$, so $X(\pi)=\sin(n\pi)=0$).

Time problem. $T'' + n^2 c^2 T = 0$ gives

$$T_n(t) = a_n\cos(nct) + b_n\sin(nct).$$

Apply the zero-initial-velocity condition $u_t(x,0)=0$: this forces $b_n = 0$ for all $n$ (since $T_n'(0) = nc\,b_n = 0$). Hence the general solution is

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

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

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

Matching coefficients term by term:

$$a_1 = 3,\qquad a_4 = -1,\qquad a_n = 0 \ (n \ne 1,4).$$

No Fourier integral is needed because the initial shape is already a finite sine series.

Explicit solution.

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

This satisfies the PDE, both fixed-end conditions, and the rest condition $u_t(x,0)=0$.

Identify the region. For $0\le x\le 2$ and $0\le y\le \sqrt{4-x^2}$, the upper bound is the circle $x^2+y^2=4$. With both $x\ge 0$ and $y\ge 0$, the region $R$ is the quarter disk of radius $2$ in the first quadrant.

Polar substitution. Let $x = r\cos\theta,\ y = r\sin\theta$, so $x^2+y^2 = r^2$ and $dA = r\,dr\,d\theta$. The region becomes

$$0 \le r \le 2,\qquad 0 \le \theta \le \frac{\pi}{2}.$$

Transform the integral.

$$\iint_R (x^2+y^2)\,dA = \int_0^{\pi/2}\int_0^2 r^2 \cdot r\,dr\,d\theta = \int_0^{\pi/2}\int_0^2 r^3\,dr\,d\theta.$$

Inner integral.

$$\int_0^2 r^3\,dr = \frac{r^4}{4}\Big|_0^2 = \frac{16}{4} = 4.$$

Outer integral.

$$\int_0^{\pi/2} 4\,d\theta = 4\cdot\frac{\pi}{2} = 2\pi.$$

Result.

$$\boxed{\int_0^2 \int_0^{\sqrt{4-x^2}} (x^2+y^2)\,dy\,dx = 2\pi \approx 6.2832}$$

(a) Gradient.

$$\nabla\phi = \left(\frac{\partial\phi}{\partial x},\,\frac{\partial\phi}{\partial y},\,\frac{\partial\phi}{\partial z}\right) = \left(2xy + z^2,\; x^2 + 2yz,\; y^2 + 2zx\right).$$

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

• $\phi_x = 2(1)(2) + (-1)^2 = 4 + 1 = 5.$
• $\phi_y = (1)^2 + 2(2)(-1) = 1 - 4 = -3.$
• $\phi_z = (2)^2 + 2(-1)(1) = 4 - 2 = 2.$

$$\nabla\phi\big|_P = 5\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}.$$

(b) Directional derivative. Unit vector along $\mathbf{a} = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}$:

$$|\mathbf{a}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3,\qquad \hat{\mathbf{a}} = \tfrac{1}{3}(2, -1, 2).$$

Then

$$D_{\hat{\mathbf{a}}}\phi = \nabla\phi\cdot\hat{\mathbf{a}} = \frac{1}{3}\big[(5)(2) + (-3)(-1) + (2)(2)\big] = \frac{1}{3}(10 + 3 + 4) = \frac{17}{3}.$$ $$D_{\hat{\mathbf{a}}}\phi = \frac{17}{3} \approx 5.667.$$

(c) Maximum rate of change. This equals $|\nabla\phi|$ and occurs in the direction of $\nabla\phi$:

$$|\nabla\phi|_P = \sqrt{5^2 + (-3)^2 + 2^2} = \sqrt{25 + 9 + 4} = \sqrt{38} \approx 6.164,$$

in the direction $\dfrac{1}{\sqrt{38}}(5\mathbf{i} - 3\mathbf{j} + 2\mathbf{k})$.

Results.

$$\boxed{\nabla\phi|_P = 5\mathbf{i} - 3\mathbf{j} + 2\mathbf{k},\quad D_{\hat{\mathbf{a}}}\phi = \tfrac{17}{3},\quad |\nabla\phi|_P = \sqrt{38}}$$

(a) Laplace transform of $t\,e^{-3t}\cos 2t$.

Start from $\mathcal{L}\{\cos 2t\} = \dfrac{s}{s^2 + 4}$.

Apply the multiplication-by-$t$ rule $\mathcal{L}\{t\,g(t)\} = -\dfrac{d}{ds}\mathcal{L}\{g(t)\}$ to $g(t)=\cos 2t$:

$$\mathcal{L}\{t\cos 2t\} = -\frac{d}{ds}\!\left(\frac{s}{s^2+4}\right) = -\frac{(s^2+4)(1) - s(2s)}{(s^2+4)^2} = -\frac{4 - s^2}{(s^2+4)^2} = \frac{s^2 - 4}{(s^2+4)^2}.$$

Now apply the first shifting theorem $\mathcal{L}\{e^{-3t}h(t)\} = H(s+3)$ with $h(t)=t\cos 2t$, i.e. replace $s$ by $s+3$:

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

Simplify $(s+3)^2 - 4 = s^2 + 6s + 5$ and $(s+3)^2 + 4 = s^2 + 6s + 13$:

$$\boxed{\mathcal{L}\{t e^{-3t}\cos 2t\} = \frac{s^2 + 6s + 5}{\left(s^2 + 6s + 13\right)^2}}$$

(b) Inverse transform of $\dfrac{2s+5}{s^2+4s+13}$.

Complete the square in the denominator: $s^2 + 4s + 13 = (s+2)^2 + 9$, where $9 = 3^2$.

Write the numerator in terms of $(s+2)$:

$$2s + 5 = 2(s+2) + 1.$$

So

$$\frac{2s+5}{(s+2)^2 + 9} = 2\cdot\frac{s+2}{(s+2)^2 + 3^2} + \frac{1}{3}\cdot\frac{3}{(s+2)^2 + 3^2}.$$

Using $\mathcal{L}^{-1}\!\left\{\dfrac{s+a}{(s+a)^2+b^2}\right\} = e^{-at}\cos bt$ and $\mathcal{L}^{-1}\!\left\{\dfrac{b}{(s+a)^2+b^2}\right\} = e^{-at}\sin bt$ with $a=2,\ b=3$:

$$\boxed{\mathcal{L}^{-1}\!\left\{\frac{2s+5}{s^2+4s+13}\right\} = e^{-2t}\left(2\cos 3t + \tfrac{1}{3}\sin 3t\right)}$$

Fourier cosine transform. With the convention

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

and $f(x)=1$ on $[0,a)$, $0$ otherwise:

$$F_c(\omega) = \sqrt{\frac{2}{\pi}}\int_0^{a} \cos(\omega x)\,dx = \sqrt{\frac{2}{\pi}}\left[\frac{\sin(\omega x)}{\omega}\right]_0^{a} = \sqrt{\frac{2}{\pi}}\,\frac{\sin a\omega}{\omega}.$$ $$\boxed{F_c(\omega) = \sqrt{\frac{2}{\pi}}\,\frac{\sin a\omega}{\omega}}$$

Cosine inversion. The inversion formula is

$$f(x) = \sqrt{\frac{2}{\pi}}\int_0^{\infty} F_c(\omega)\cos(\omega x)\,d\omega = \frac{2}{\pi}\int_0^{\infty} \frac{\sin a\omega}{\omega}\cos(\omega x)\,d\omega.$$

Evaluate at $x = 0$. Here $f(0) = 1$ (interior point of the interval $[0,a)$), so

$$1 = \frac{2}{\pi}\int_0^{\infty} \frac{\sin a\omega}{\omega}\,d\omega.$$

Solving,

$$\int_0^{\infty} \frac{\sin a\omega}{\omega}\,d\omega = \frac{\pi}{2}.$$

(For $a>0$ this is the standard Dirichlet integral; the value is independent of $a$, since substituting $u=a\omega$ gives $\int_0^\infty \frac{\sin u}{u}\,du$.)

Result.

$$\boxed{\int_0^{\infty} \frac{\sin a\omega}{\omega}\,d\omega = \frac{\pi}{2}\quad (a>0)}$$

Check harmonicity. Compute second partials of $u = x^3 - 3xy^2 + 2y$:

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

Then

$$u_{xx} + u_{yy} = 6x + (-6x) = 0,$$

so $u$ satisfies Laplace's equation and is harmonic.

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

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

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

where $g(x)$ is an unknown function of $x$.

Differentiate with respect to $x$: $v_x = 6xy + g'(x)$. The second CR equation requires $v_x = -u_y = -(-6xy + 2) = 6xy - 2$. Hence

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

Taking the constant $C=0$:

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

Form the analytic function.

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

Group the cubic and linear parts. Note $x^3 - 3xy^2 + i(3x^2 y - y^3) = (x+iy)^3 = z^3$. The remaining terms: $2y - 2ix = -2i(x + iy) = -2iz$. Therefore

$$f(z) = z^3 - 2iz \;(+\,iC).$$

Result.

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

(a) Classification. For a second-order PDE

$$A u_{xx} + B u_{xy} + C u_{yy} + (\text{lower-order}) = 0,$$

the type is determined by the discriminant $B^2 - 4AC$:

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

Here $A = 3,\ B = 2,\ C = 5$:

$$B^2 - 4AC = (2)^2 - 4(3)(5) = 4 - 60 = -56 < 0.$$

Since the discriminant is negative, the equation is elliptic (everywhere, as the coefficients are constant).

(b) Eliminate $f$ from $z = f(x^2 - y^2)$.

Let $w = x^2 - y^2$, so $z = f(w)$. Differentiate:

$$p = \frac{\partial z}{\partial x} = f'(w)\cdot\frac{\partial w}{\partial x} = f'(w)\cdot(2x),$$ $$q = \frac{\partial z}{\partial y} = f'(w)\cdot\frac{\partial w}{\partial y} = f'(w)\cdot(-2y).$$

Divide to eliminate the unknown $f'(w)$ (assuming it is nonzero):

$$\frac{p}{q} = \frac{2x}{-2y} = -\frac{x}{y}.$$

Cross-multiplying:

$$p\,y = -q\,x \;\Rightarrow\; y\,p + x\,q = 0.$$

Result.

$$\boxed{(a)\ \text{Elliptic}\ (B^2-4AC=-56<0);\qquad (b)\ y\,\frac{\partial z}{\partial x} + x\,\frac{\partial z}{\partial y} = 0}$$