Engineering Mathematics-I

Using the Beta function, $B(3,4)=\frac{\Gamma(3)\Gamma(4)}{\Gamma(7)}=\frac{2!\,3!}{6!}=\boxed{\frac{1}{60}}$.

For $y=f(x)$, revolved from $x=a$ to $x=b$, the surface area is $\displaystyle S=2\pi\int_a^b y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$.

A sequence $\{a_n\}$ converges to $L$ if $\displaystyle\lim_{n\to\infty}a_n=L$, where $L$ is finite.

Let $a_n=\frac1{n(n+1)}=\frac1n-\frac1{n+1}$. Therefore, $\displaystyle\sum_{n=2}^{\infty}a_n=\frac12$ by telescoping. The series converges.

If $f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos nx+b_n\sin nx)$, then $\displaystyle\frac1\pi\int_{-\pi}^{\pi}f^2(x)\,dx=\frac{a_0^2}{2}+\sum_{n=1}^{\infty}(a_n^2+b_n^2)$.

$\displaystyle a_0=\frac1\pi\int_{-\pi}^{\pi}|x|\,dx=\frac2\pi\int_0^\pi x\,dx=\boxed{\pi}$.

At a local minimum, $f_x=f_y=0$ and the Hessian is positive semidefinite. The usual strict-minimum test is $D=f_{xx}f_{yy}-f_{xy}^2>0$ and $f_{xx}>0$.

$f_x=2xy^2+y^2$, $f_y=2x^2y+2xy$, $f_z=-2z$. At $(3,1,1)$, $\boxed{\nabla f=7\mathbf i+24\mathbf j-2\mathbf k}$.

$\displaystyle\int_0^b\left[y\frac{x^2}{2}\right]_0^y dy=\frac12\int_0^b y^3\,dy=\boxed{\frac{b^4}{8}}$.

For a closed surface $S$ enclosing volume $V$, $\displaystyle\iint_S\mathbf F\cdot\mathbf n\,dS=\iiint_V(\nabla\cdot\mathbf F)\,dV$, where $\mathbf n$ is the outward unit normal.

Split the integral:

$I=\int_0^\infty\frac{x^2}{(1+x)^{10}}dx+\int_0^\infty\frac{x^6}{(1+x)^{10}}dx.$

Using $\displaystyle\int_0^\infty\frac{x^{p-1}}{(1+x)^{p+q}}dx=B(p,q)$,

$I=B(3,7)+B(7,3)=2\frac{\Gamma(3)\Gamma(7)}{\Gamma(10)}=\boxed{\frac1{126}}.$

Put $h=x-\frac\pi2$. Then $\sin x=\cos h$. Hence

$\boxed{\sin x=1-\frac{(x-\pi/2)^2}{2!}+\frac{(x-\pi/2)^4}{4!}-\frac{(x-\pi/2)^6}{6!}+\cdots}$

Since $x^2$ is even, $b_n=0$. Its Fourier coefficients give

$\boxed{x^2=\frac{\pi^2}{3}+4\sum_{n=1}^{\infty}\frac{(-1)^n\cos nx}{n^2}}.$

Parseval's identity gives

$\frac1\pi\int_{-\pi}^{\pi}x^4\,dx=\frac12\left(\frac{2\pi^2}{3}\right)^2+16\sum_{n=1}^{\infty}\frac1{n^4}.$

Since the integral term equals $2\pi^4/5$,

$16\sum_{n=1}^{\infty}\frac1{n^4}=\frac{8\pi^4}{45},\qquad\boxed{\sum_{n=1}^{\infty}\frac1{n^4}=\frac{\pi^4}{90}}.$

Let $u=e^{xyz}$. First, $u_x=yz e^{xyz}$. Differentiating successively with respect to $y$ and $z$ gives

$u_{xy}=z(1+xyz)e^{xyz},$

$\boxed{u_{xyz}=(1+3xyz+x^2y^2z^2)e^{xyz}}.$

Incompressibility requires $\nabla\cdot\mathbf V=0$. Here

$\nabla\cdot\mathbf V=\frac{\partial(y+z)}{\partial x}+\frac{\partial(z+x)}{\partial y}+\frac{\partial(x+y)}{\partial z}=0+0+0=0.$

Therefore, the flow is incompressible.

The region is $0\le x\le1$, $e^x\le y\le e$. Reversing the order gives $1\le y\le e$ and $0\le x\le\ln y$.

$I=\int_1^e\int_0^{\ln y}\frac1{\ln y}\,dx\,dy=\int_1^e1\,dy=\boxed{e-1}.$

Integrate first with respect to $x$:

$\int_0^{yz}xyz\,dx=\frac{y^3z^3}{2}.$

Then

$I=\int_1^2\int_1^z\frac{y^3z^3}{2}\,dy\,dz=\frac18\int_1^2(z^7-z^3)\,dz.$

$I=\left[\frac{z^8}{64}-\frac{z^4}{32}\right]_1^2=\boxed{\frac{225}{64}}.$

(a) Put $t=x^6$. Then

$I=\frac16\int_0^\infty\frac{t^{-2/3}}{1+t}\,dt=\frac16B\left(\frac13,\frac23\right)=\frac16\Gamma\left(\frac13\right)\Gamma\left(\frac23\right)=\boxed{\frac{\pi}{3\sqrt3}}.$

(b)

$J=\int_0^1x^{-1/2}(1-x)^{1/2}\,dx=B\left(\frac12,\frac32\right)=\frac{\Gamma(1/2)\Gamma(3/2)}{\Gamma(2)}=\boxed{\frac\pi2}.$

The function is odd, so $a_0=a_n=0$. The sine coefficients are

$b_n=\frac2\pi\int_0^\pi\sin(nx)\,dx=\frac{2(1-(-1)^n)}{\pi n}.$

Thus only odd terms remain:

$\boxed{f(x)=\frac4\pi\left(\sin x+\frac{\sin3x}{3}+\frac{\sin5x}{5}+\cdots\right)}.$

Set $x=\pi/2$ to obtain $1=\frac4\pi(1-\frac13+\frac15-\cdots)$, hence the required result.

Minimize/maximize $D^2=(x-3)^2+(y-4)^2+(z-12)^2$ subject to $x^2+y^2+z^2=1$. Lagrange multipliers show the extremal sphere point lies on the line through the origin and $(3,4,12)$.

The given point has distance $\sqrt{3^2+4^2+12^2}=13$ from the origin. The sphere has radius $1$, so the nearest and farthest distances are

$\boxed{D_{\min}=13-1=12,\qquad D_{\max}=13+1=14}.$

They occur at $(3/13,4/13,12/13)$ and $(-3/13,-4/13,-12/13)$ respectively.

Since $r=\sqrt{x^2+y^2}$,

$u_x=f'(r)\frac{x}{r},\qquad u_y=f'(r)\frac{y}{r}.$

Differentiating again,

$u_{xx}=f''(r)\frac{x^2}{r^2}+f'(r)\left(\frac1r-\frac{x^2}{r^3}\right),$

$u_{yy}=f''(r)\frac{y^2}{r^2}+f'(r)\left(\frac1r-\frac{y^2}{r^3}\right).$

Adding and using $x^2+y^2=r^2$ yields

$\boxed{u_{xx}+u_{yy}=f''(r)+\frac{f'(r)}r}.$

Take $P=xy+y^2$ and $Q=x^2$. The curves meet at $x=0,1$; the region is $0\le x\le1$, $x^2\le y\le x$.

Double integral: Green's theorem gives

$\iint_R(Q_x-P_y)\,dA=\int_0^1\int_{x^2}^{x}(2x-x-2y)\,dy\,dx$

$=\int_0^1\int_{x^2}^{x}(x-2y)\,dy\,dx=\int_0^1(-x^3+x^4)\,dx=-\frac1{20}.$

Line integral: Orient $C$ positively. Along $y=x^2$, $(x,y):(0,0)\to(1,1)$, the integral is $19/20$. Along $y=x$, $(1,1)\to(0,0)$, it is $-1$. Their sum is $19/20-1=-1/20$.

Both sides equal $\boxed{-1/20}$, so Green's theorem is verified.