(ii) Find the volume of the solid obtained by revolving one arc of the cycloid $x = a(\theta + \sin \theta)$ and $y = a(1 + \cos \theta)$ about the x-axis.

1. Parametric Equations: The given parametric equations are:

   $x = a(\theta + \sin \theta)$

   $y = a(1 + \cos \theta)$

2. Volume of Revolution:

   The volume V of a solid of revolution about the x-axis using the disk method is given by:

   $V = \pi \int_{a}^{b} y^2 \frac{dx}{d\theta} \, d\theta$

3. Compute $\frac{dx}{d\theta}$:

   First, we need to find $\frac{dx}{d\theta}$:

   $x = a(\theta + \sin \theta)$

   $\frac{dx}{d\theta} = a(1 + \cos \theta)$

4. Square of y:

   Next, we find $y^2$:

   $y = a(1 + \cos \theta)$

   $y^2 = a^2(1 + \cos \theta)^2$

5. Limits of Integration:

   For one arc of the cycloid, $\theta$ ranges from 0 to $2\pi$.

6. Set Up the Integral:

   Substitute $y^2$ and $\frac{dx}{d\theta}$ into the integral:

   $V = \pi \int_{0}^{2\pi} [a^2 (1 + \cos \theta)^2] [a (1 + \cos \theta)] \, d\theta$

   $V = \pi a^3 \int_{0}^{2\pi} (1 + \cos \theta)^3 \, d\theta$

7. Expand and Integrate: Expand $(1 + \cos \theta)^3$:

   $(1 + \cos \theta)^3 = 1 + 3\cos \theta + 3\cos^2 \theta + \cos^3 \theta$

   Use the identities $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$ and $\cos^3 \theta = \frac{3\cos \theta + \cos 3\theta}{4}$:

   $(1 + \cos \theta)^3 = 1 + 3\cos \theta + 3\left(\frac{1 + \cos 2\theta}{2}\right) + \left(\frac{3\cos \theta + \cos 3\theta}{4}\right)$

   Simplify:

   $(1 + \cos \theta)^3 = 1 + 3\cos \theta + \frac{3}{2}(1 + \cos 2\theta) + \frac{3}{4}\cos \theta + \frac{1}{4}\cos 3\theta$

   $(1 + \cos \theta)^3 = \frac{5}{2} + \frac{15}{4}\cos \theta + \frac{3}{2}\cos 2\theta + \frac{1}{4}\cos 3\theta$

   Integrate each term:

   $\int_{0}^{2\pi} \cdot$

   $\left(\frac{5}{2} + \frac{15}{4}\cos \theta + \frac{3}{2}\cos 2\theta + \frac{1}{4}\cos 3\theta\right) \cdot$

   $d\theta$

   Since the integrals of $\cos \theta$, $\cos 2\theta$, and $\cos 3\theta$ over one period are zero, we only need to integrate the constant term:

   $\int_{0}^{2\pi} \frac{5}{2} \, d\theta = \frac{5}{2} \theta \Big|_{0}^{2\pi} = \frac{5}{2} \cdot 2\pi = 5\pi$

8. Final Volume:

   $V = \pi a^3 \cdot 5\pi = 5\pi^2 a^3$

   Thus, the volume of the solid is: $\boxed{5\pi^2 a^3}$.

5. (i) In a plane triangle, find the maximum value of cos A cos B cos C.

We have $A + B + C = \pi$ so that $C = \pi - (A + B)$.

$\cos A \cos B \cos C = \cos A \cos B \cos \left[\pi - (A + B)\right]$

$= -\cos A \cos B \cos (A + B) = f(A, B), \text{ say.}$

We get,

$\frac{\partial f}{\partial A} = \cos B \left[\sin A \cos (A + B) + \right. \\$

$\left. \\ \cos A \sin (A + B)\right]$

$= \cos B \sin (2A + B)$

and

$\frac{\partial f}{\partial B} = \cos A \sin (A + 2B)$

$\frac{\partial f}{\partial A} = 0, \quad \frac{\partial f}{\partial B} = 0 \quad \text{only when } A = B = \pi/3.$

Also

$r = \frac{\partial^2 f}{\partial A^2} = 2 \cos B \cos (2A + B), \quad t = \frac{\partial^2 f}{\partial B^2} = 2 \cos A \cos (A + 2B)$

$s = \frac{\partial^2 f}{\partial A \partial B} = -\sin B \sin (2A + B) + \cos B \cos (2A + B) = \cos (2A + 2B)$

When $A = B = \pi/3$ , $r = -1$, $s = -1/2, t = -1$ so that $rt - s^2 = 3/4$.

These show that f(A, B) is maximum for $A = B = \pi/3$.

Then $C = \pi - (A + B) = \pi/3$.

Hence $\cos A \cos B \cos C$ is maximum when each of the angles is $\pi/3$ i.e., the triangle is equilateral and its maximum value is $1/8$.

(ii) If $u = \log(x^3 + y^3 + z^3 - 3xyz)$, show that:

$\left( \frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z} \right)^2 u = \frac{-9}{(x + y + z)^2}$

We have,

$\frac{\partial u}{\partial x} = \frac{3x^2 - 3yz}{x^3 + y^3 + z^3 - 3xyz}, \quad \frac{\partial u}{\partial y} = \frac{3y^2 - 3zx}{x^3 + y^3 + z^3 - 3xyz}, \quad \frac{\partial u}{\partial z} = \frac{3z^2 - 3xy}{x^3 + y^3 + z^3 - 3xyz}$

$\therefore \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = \frac{3(x^2 + y^2 + z^2 - xy - yz - zx)}{x^3 + y^3 + z^3 - 3xyz}$

$= \frac{3(x^2 + y^2 + z^2 - xy - yz - zx)}{(x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)} = \frac{3}{x + y + z}$

$\text{Now} \quad \left(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}\right)^2 u = \left(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}\right) \left(\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z}\right)$

$= \left(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}\right) \left(\frac{3}{x + y + z}\right)$

$= - \frac{3}{(x + y + z)^2} - \frac{3}{(x + y + z)^2} - \frac{3}{(x + y + z)^2} = - \frac{9}{(x + y + z)^2}.$

6. (i) If $u = x^y$, show that $\frac{\partial^2 u}{\partial x^2 \partial y} = \frac{\partial^3 u}{\partial x \partial y \partial z}$.

Given:

$u = x^y$

We want to show that:

$\frac{\partial^2 u}{\partial x^2 \partial y} = \frac{\partial^3 u}{\partial x \partial y \partial z}$

First, we compute the second partial derivative $\frac{\partial^2 u}{\partial x^2 \partial y}$:

$\frac{\partial u}{\partial x} = y x^{y-1}$

$\frac{\partial^2 u}{\partial x^2} = y (y-1) x^{y-2}$

$\frac{\partial^2 u}{\partial x^2 \partial y} = \frac{\partial}{\partial y} \left( y (y-1) x^{y-2} \right)$

Next, we compute the third partial derivative $\frac{\partial^3 u}{\partial x \partial y \partial z}$:

Since $u = x^y$ does not depend on z, we have:

$\frac{\partial^3 u}{\partial x \partial y \partial z} = \frac{\partial}{\partial z} \left( \frac{\partial^2 u}{\partial x \partial y} \right) = 0$

Similarly, for $\frac{\partial^2 u}{\partial x^2 \partial y}$, we have:

$\frac{\partial^2 u}{\partial x^2 \partial y} = 0$

Therefore,

$\boxed{\frac{\partial^2 u}{\partial x^2 \partial y} = \frac{\partial^3 u}{\partial x \partial y \partial z}}$

(ii) Show that the rectangular solid of maximum volume that can be inscribed in a sphere is a cube.

Let 2x, 2y, 2z be the length, breadth, and height of the rectangular solid so that its volume V = 8xyz.

$V = 8xyz \quad \text{...(i)}$

Let R be the radius of the sphere so that $x^2 + y^2 + z^2 = R^2$.

$F(x, y, z) = 8xyz + \lambda (x^2 + y^2 + z^2 - R^2) \quad \text{...(ii)}$

and

$\frac{\partial F}{\partial x} = 0, \, \frac{\partial F}{\partial y} = 0, \, \frac{\partial F}{\partial z} = 0$

give

$8yz + 2x\lambda = 0, \quad 8xz + 2y\lambda = 0, \quad 8xy + 2z\lambda = 0$

or

$2x^2\lambda = -8xyz = 2y^2\lambda = 2z^2\lambda$

Thus, for a maximum volume $x = y = z$.

i.e., the rectangular solid is a cube.