[In this question, all measurements of length are in metres.] A monument is to be constructed which comprises two parts – the main structure and the base. The main structure is in the form of the solid obtained when the region $R$ bounded by the curves $y = 3 \sin\left(2x^2\right)$ and $y = x^2$, and the line $x = \sqrt{\frac{\pi}{8}}$ is rotated by $2\pi$ radians about the $y$-axis. (See Figure 1.) The main structure is to be mounted on top of a base which is a solid of uniform thickness. The horizontal cross-sectional area of the base is in the shape of region $Q$, bounded by the curve $y = \frac{8\sqrt{x}}{1+x^3}$ and the curve $y = 4x^{\frac{7}{2}}$. (See Figure 2.)

1. Set up the integration by parts.

Let $u = \sin^{-1}\left(\frac{t}{3}\right)$ and $\frac{\mathrm{d}v}{\mathrm{d}t} = 1$.

Then, differentiating $u$ with respect to $t$ using the chain rule:
$\frac{\mathrm{d}u}{\mathrm{d}t} = \frac{1}{\sqrt{1 - \left(\frac{t}{3}\right)^2}} \cdot \frac{1}{3} = \frac{1}{\sqrt{\frac{9-t^2}{9}}} \cdot \frac{1}{3} = \frac{3}{\sqrt{9-t^2}} \cdot \frac{1}{3} = \frac{1}{\sqrt{9-t^2}}$

Integrating $\frac{\mathrm{d}v}{\mathrm{d}t}$ with respect to $t$:
$v = t$

Applying the integration by parts formula $\int u \frac{\mathrm{d}v}{\mathrm{d}t} \, \mathrm{d}t = uv - \int v \frac{\mathrm{d}u}{\mathrm{d}t} \, \mathrm{d}t$:
$\int \sin^{-1}\left(\frac{t}{3}\right) \; \mathrm{d}t = t \sin^{-1}\left(\frac{t}{3}\right) - \int \frac{t}{\sqrt{9 - t^2}} \; \mathrm{d}t$

2. Integrate the remaining algebraic term.

Notice that the derivative of the inner function $9 - t^2$ is $-2t$. We manipulate the integral to use the standard form $\int \mathrm{f}'(t)[\mathrm{f}(t)]^n \, \mathrm{d}t$:

$\int \frac{t}{\sqrt{9 - t^2}} \; \mathrm{d}t = -\frac{1}{2} \int -2t (9 - t^2)^{-\frac{1}{2}} \; \mathrm{d}t$
$= -\frac{1}{2} \left[ \frac{(9 - t^2)^{\frac{1}{2}}}{\frac{1}{2}} \right] = -\sqrt{9 - t^2}$

Substituting this back into the integration by parts result:
$\int \sin^{-1}\left(\frac{t}{3}\right) \; \mathrm{d}t = t \sin^{-1}\left(\frac{t}{3}\right) + \sqrt{9 - t^2} + C$

3. Evaluate the definite integral with the given limits $0$ to $p$.

$\int_{0}^{p} \sin^{-1}\left(\frac{t}{3}\right) \; \mathrm{d}t = \left[ t \sin^{-1}\left(\frac{t}{3}\right) + \sqrt{9 - t^2} \right]_{0}^{p}$
$= \left( p \sin^{-1}\left(\frac{p}{3}\right) + \sqrt{9 - p^2} \right) - \left( 0 \cdot \sin^{-1}(0) + \sqrt{9 - 0^2} \right)$
$= p \sin^{-1}\left(\frac{p}{3}\right) + \sqrt{9 - p^2} - \sqrt{9}$
$= p \sin^{-1}\left(\frac{p}{3}\right) + \sqrt{9 - p^2} - 3 \quad \text{(Shown)}$

1. Express $x^2$ in terms of $y$ for the boundary curves to prepare for integration with respect to $y$.

For the upper curve:
$y = 3\sin(2x^2) \implies \frac{y}{3} = \sin(2x^2) \implies 2x^2 = \sin^{-1}\left(\frac{y}{3}\right) \implies x^2 = \frac{1}{2}\sin^{-1}\left(\frac{y}{3}\right)$

For the lower curve:
$y = x^2 \implies x^2 = y$

2. Determine the $y$-coordinates of the region's boundaries.

At the right boundary $x = \sqrt{\frac{\pi}{8}}$, the $y$-coordinates are:
Lower curve: $y = \left(\sqrt{\frac{\pi}{8}}\right)^2 = \frac{\pi}{8}$
Upper curve: $y = 3\sin\left(2\left(\sqrt{\frac{\pi}{8}}\right)^2\right) = 3\sin\left(\frac{\pi}{4}\right) = \frac{3}{\sqrt{2}}$

3. Set up the definite integral for the exact volume using the disc method.

The region is rotated about the $y$-axis, so $V = \pi \int (x_R^2 - x_L^2) \, \mathrm{d}y$.
The left boundary is the upper curve $x_L^2 = \frac{1}{2}\sin^{-1}\left(\frac{y}{3}\right)$ for $0 \le y \le \frac{3}{\sqrt{2}}$.
The right boundary changes along the region: it is the lower curve $x_R^2 = y$ for $0 \le y \le \frac{\pi}{8}$, and the vertical line $x_R^2 = \frac{\pi}{8}$ for $\frac{\pi}{8} \le y \le \frac{3}{\sqrt{2}}$.

$V = \pi \int_{0}^{\frac{\pi}{8}} \left( y - \frac{1}{2}\sin^{-1}\left(\frac{y}{3}\right) \right) \mathrm{d}y + \pi \int_{\frac{\pi}{8}}^{\frac{3}{\sqrt{2}}} \left( \frac{\pi}{8} - \frac{1}{2}\sin^{-1}\left(\frac{y}{3}\right) \right) \mathrm{d}y$

Separating the outer and inner volume integrals:
$V = \pi \left( \int_{0}^{\frac{\pi}{8}} y \, \mathrm{d}y + \int_{\frac{\pi}{8}}^{\frac{3}{\sqrt{2}}} \frac{\pi}{8} \, \mathrm{d}y \right) - \frac{\pi}{2} \int_{0}^{\frac{3}{\sqrt{2}}} \sin^{-1}\left(\frac{y}{3}\right) \mathrm{d}y$

4. Evaluate the polynomial integrals for the outer volume component.

$\int_{0}^{\frac{\pi}{8}} y \, \mathrm{d}y = \left[ \frac{y^2}{2} \right]_{0}^{\frac{\pi}{8}} = \frac{\pi^2}{128}$
$\int_{\frac{\pi}{8}}^{\frac{3}{\sqrt{2}}} \frac{\pi}{8} \, \mathrm{d}y = \frac{\pi}{8} \left( \frac{3}{\sqrt{2}} - \frac{\pi}{8} \right) = \frac{3\sqrt{2}\pi}{16} - \frac{\pi^2}{64}$

Outer Volume $= \pi \left( \frac{\pi^2}{128} + \frac{3\sqrt{2}\pi}{16} - \frac{\pi^2}{64} \right) = \frac{3\sqrt{2}\pi^2}{16} - \frac{\pi^3}{128}$

5. Apply the result from part (a) to evaluate the inner volume and find the final exact volume.

Using the given result from part (a) with $p = \frac{3}{\sqrt{2}}$:
$\int_{0}^{\frac{3}{\sqrt{2}}} \sin^{-1}\left(\frac{t}{3}\right) \mathrm{d}t = \frac{3}{\sqrt{2}} \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) + \sqrt{9 - \left(\frac{3}{\sqrt{2}}\right)^2} - 3$
$= \frac{3}{\sqrt{2}} \left(\frac{\pi}{4}\right) + \sqrt{4.5} - 3 = \frac{3\sqrt{2}\pi}{8} + \frac{3\sqrt{2}}{2} - 3$

Inner Volume $= \frac{\pi}{2} \left( \frac{3\sqrt{2}\pi}{8} + \frac{3\sqrt{2}}{2} - 3 \right) = \frac{3\sqrt{2}\pi^2}{16} + \frac{3\sqrt{2}\pi}{4} - \frac{3\pi}{2}$

Subtracting the Inner Volume from the Outer Volume:
$V = \left( \frac{3\sqrt{2}\pi^2}{16} - \frac{\pi^3}{128} \right) - \left( \frac{3\sqrt{2}\pi^2}{16} + \frac{3\sqrt{2}\pi}{4} - \frac{3\pi}{2} \right)$
$V = \frac{3\pi}{2} - \frac{3\sqrt{2}\pi}{4} - \frac{\pi^3}{128}$

1. Find the $x$-coordinate of the non-zero point of intersection of the two curves.

$\frac{8\sqrt{x}}{1+x^3} = 4x^{\frac{7}{2}}$
$2\sqrt{x} = x^{\frac{7}{2}}(1+x^3)$
For $x > 0$, divide both sides by $\sqrt{x}$ (which is $x^{\frac{1}{2}}$):
$2 = x^3(1+x^3)$
$x^6 + x^3 - 2 = 0$
$(x^3 - 1)(x^3 + 2) = 0$
Since $x$ is real and positive in the first quadrant, $x^3 = 1 \implies x = 1$.

2. Set up the definite integral for the area of region $Q$.

From Figure 2, the curve $y = \frac{8\sqrt{x}}{1+x^3}$ lies above the curve $y = 4x^{\frac{7}{2}}$ for $0 \le x \le 1$.
$\text{Area} = \int_{0}^{1} \left( \frac{8\sqrt{x}}{1+x^3} - 4x^{\frac{7}{2}} \right) \mathrm{d}x$

3. Evaluate the integral and round to 1 decimal place.

Using a graphing calculator to evaluate the definite integral directly:
$\text{Area} \approx 3.2999 \text{ m}^2$
$\text{Area} = 3.3 \text{ m}^2 \text{ (correct to 1 d.p.)}$

(Note: Evaluating exactly via the substitution $u = x^{\frac{3}{2}}$ yields $\left[ \frac{16}{3}\tan^{-1}\left(x^{\frac{3}{2}}\right) - \frac{8}{9}x^{\frac{9}{2}} \right]_0^1 = \frac{4\pi}{3} - \frac{8}{9} \approx 3.3$)

1. Calculate the exact volume of the main structure (from part b).

Using the washer method for region $R$ rotated about the $y$-axis, the boundaries are $x = \sqrt{\frac{1}{2}\sin^{-1}\left(\frac{y}{3}\right)}$ and $x = \sqrt{y}$ for $y \in [0, \frac{\pi}{8}]$, and $x = \sqrt{\frac{\pi}{8}}$ for $y \in [\frac{\pi}{8}, \frac{3}{\sqrt{2}}]$.
$V_R = \int_0^{\frac{\pi}{8}} \pi \left( y - \frac{1}{2}\sin^{-1}\left(\frac{y}{3}\right) \right) \mathrm{d}y + \int_{\frac{\pi}{8}}^{\frac{3}{\sqrt{2}}} \pi \left( \frac{\pi}{8} - \frac{1}{2}\sin^{-1}\left(\frac{y}{3}\right) \right) \mathrm{d}y$
Using the integral result from (a), this evaluates to:
$V_R = \frac{3\pi}{2} - \frac{3\pi\sqrt{2}}{4} - \frac{\pi^3}{128} \approx 1.13799 \text{ m}^3$

2. Calculate the exact area of the base, region $Q$ (from part c).

Find the intersection point of the curves:
$\frac{8\sqrt{x}}{1+x^3} = 4x^{\frac{7}{2}} \implies 2 = x^3(1+x^3) \implies x = 1$
$A_Q = \int_0^1 \left( \frac{8\sqrt{x}}{1+x^3} - 4x^{\frac{7}{2}} \right) \mathrm{d}x$
Using the substitution $u = x^{\frac{3}{2}}$ for the first term:
$A_Q = \frac{16}{3} \big[ \tan^{-1} u \big]_0^1 - \left[ \frac{8}{9}x^{\frac{9}{2}} \right]_0^1 = \frac{4\pi}{3} - \frac{8}{9} \approx 3.29990 \text{ m}^2$

3. Determine the mass and total weight of the monument assuming the maximum claimed base thickness of 70 cm.

Mass of granite main structure:
$1.13799 \times 1463.46 \approx 1665.4 \text{ kg}$
Weight of main structure:
$\frac{1665.4}{1000} \times 9.81 \approx 16.338 \text{ kN}$
For a base thickness of $h = 0.70 \text{ m}$, the volume of the marble base is:
$V_{\text{base}} = A_Q \times 0.70 \approx 3.29990 \times 0.70 \approx 2.30993 \text{ m}^3$
Mass of marble base:
$2.30993 \times 2550 \approx 5890.3 \text{ kg}$
Weight of marble base:
$\frac{5890.3}{1000} \times 9.81 \approx 57.784 \text{ kN}$
Total weight of the monument:
$W_{\text{total}} = 16.338 + 57.784 = 74.122 \text{ kN}$

4. Calculate the weight per square metre and conclude on the engineer's claim.

The weight per square metre on the ground is the total weight divided by the base contact area $A_Q$:
$\text{Pressure} = \frac{74.122}{3.29990} \approx 22.46 \text{ kN/m}^2$
Since $22.46 \text{ kN/m}^2 > 20 \text{ kN/m}^2$, the total weight per square metre exceeds the allowable limit when the thickness is $70 \text{ cm}$. Thus, the engineer's claim is inaccurate.