Fishy Volume

The volume of revolution of the fish's "body" is $\begin{aligned} V_0 & = \; \iint_{x,y \ge 0} 2\pi y\,dx\,dy \; = \; 2\pi\int_0^{\frac14\pi} \int_0^{\frac{\sin4\theta}{\sin\theta}}r\sin\theta\,r\,dr\,d\theta \\ & = \; \tfrac23\pi\int_0^{\frac14\pi} \left(\tfrac{\sin4\theta}{\sin\theta}\right)^3 \sin\theta\,d\theta \; = \; \tfrac23\pi\int_0^{\frac14\pi} \frac{\sin^34\theta}{\sin^2\theta}\,d\theta \\ & = \; \tfrac{128}{3}\pi \int_0^{\frac14\pi} \cos^32\theta\cos^3\theta \sin\theta\,d\theta \; = \; \tfrac{128}{3}\pi\int_{\frac{1}{\sqrt{2}}}^1 (2x^2 - 1)^3x^3\,dx \\ & = \; \tfrac{12}{5}\pi \end{aligned}$ and, similarly, the volume of revolution of the "tail" is $\begin{aligned} V_1 & = \; -2\pi\int_{\frac14\pi}^{\frac12\pi} \int_0^{\frac{\sin4\theta}{\sin\theta}}r\sin\theta\,r\,dr\,d\theta \;=\; -\tfrac{128}{3}\pi\int_0^{\frac{1}{\sqrt{2}}} (2x^2 - 1)^3x^3\,dx \\ & = \; \tfrac{4}{15}\pi \end{aligned}$ and so the total volume of revolution is $V_0 + V_1 = \tfrac83\pi$ , making the answer $8 + 3 = \boxed{11}$ .

Using the double angle identities, $r = \frac{\sin 4\theta}{\sin \theta}$ converts to $r = 4 \cos \theta (\cos^2 \theta - \sin^2 \theta)$ , using $x = r \cos \theta$ and $y = r \sin \theta$ , it further converts to $r = 4 \frac{x}{r} (\frac{x^2}{r^2} - \frac{y^2}{r^2})$ or $r^4 = 4x^3 - 4xy^2$ , and using $r^2 = x^2 + y^2$ , it converts to $(x^2 + y^2)^2 = 4x^3 - 4xy^2$ , which can be rearranged to $y = \pm \sqrt{\pm 2x\sqrt{2x +1} - x^2 - 2x}$ .

Now $y = \sqrt{2x\sqrt{2x +1} - x^2 - 2x}$ has real solutions from $-\frac{1}{2} \leq x \leq 4$ (for the body and tail of the fish) and $y = \sqrt{-2x\sqrt{2x +1} - x^2 - 2x}$ has real solutions from $-\frac{1}{2} \leq x \leq 0$ (for the tail of the fish).

A rotation around the $x$ -axis for the body of the fish gives cross-sections of circles, and so its volume is $V_{body}$ $=$ $\int_{0}^{4} \pi r^2 dx$ $=$ $\pi \int_{0}^{4} (2x\sqrt{2x +1} - x^2 - 2x) dx$ $=$ $\pi(\frac{2}{3}x\sqrt{2x + 1}^3 - \frac{2}{15} \sqrt{2x + 1}^5 - \frac{1}{3}x^3 - x^2)]_0 ^4$ $=$ $\frac{12}{5}\pi$ .

A rotation around the $x$ -axis for the tail of the fish gives cross-sections of washers, and so its volume is $V_{tail}$ $=$ $\int_{-\frac{1}{2}}^{0} (\pi R^2 - \pi r^2) dx$ $=$ $\pi \int_{-\frac{1}{2}}^{0} ((-2x\sqrt{2x +1} - x^2 - 2x) - (2x\sqrt{2x +1} - x^2 - 2x)) dx$ $=$ $\pi \int_{-\frac{1}{2}}^{0} -4x\sqrt{2x +1} dx$ $=$ $\pi(-\frac{4}{3}x\sqrt{2x + 1}^3 + \frac{4}{15} \sqrt{2x + 1}^5)]_{-\frac{1}{2}} ^0$ $=$ $\frac{4}{15}\pi$ .

The total volume of the fish is $V = V_{body} + V_{tail}$ $=$ $\frac{12}{5}\pi + \frac{4}{15}\pi$ $=$ $\frac{8}{3}\pi$ , so $a = 8$ and $b = 3$ , and $a + b = \boxed{11}$ .

A less elegant approach:

Clearly $y=r\cdot \sin(\theta )$ and $x=r\cdot \cos(\theta )$ , and since solving $r=\tfrac{\sin(4\theta )}{\sin(\theta )}=0,\;\; 0< \theta <\tfrac{\pi}{2}$ yields $\theta=\tfrac{\pi }{4}$ , the volume of revolution of the fish's body can be given by $\begin{aligned} V_{\text{body}}&=\int_{x\; :\; \theta =\frac{\pi }{4}}^{x\; :\; \theta =0}\pi \cdot y^2\; dx \\ &=\int_{x\; :\; \theta =\frac{\pi }{4}}^{x\; :\; \theta =0}\pi \cdot \sin^2(4\theta )\cdot \dfrac{d\theta }{dx}\; dx \\ &=\int_{\frac{\pi }{4}}^0\pi \cdot \sin^2(4\theta )\cdot \left(\dfrac{2\sin(6\theta )-2\sin(4\theta )-2\sin(2\theta )}{1-\cos(2\theta )}\right)\; d\theta \\ &=\dfrac{12\pi }{5}. \end{aligned}$ To find the volume of revolution of the fish's tail, we first determine the value of $\theta$ in the range $\tfrac{\pi }{4}<\theta <\tfrac{\pi }{2}$ for which the value of $y'$ (with respect to $x$ ) is infinite. This can be done by maximising $x$ through solving the equation $\tfrac{d}{d\theta }\left(r\cdot \cos(\theta )\right)=0,\;\; \tfrac{\pi }{4}<\theta <\tfrac{\pi }{2}$ , which yields $\theta =\tfrac{\pi }{3}$ . So we split the tail at this value of $\theta$ , making the volume of revolution of the fish's tail equal to $\begin{aligned} V_{\text{tail}} &= \int_{x\; :\; \theta = \frac{\pi }{3}}^{x\; :\; \theta = \frac{\pi }{2}}\pi \cdot \sin^2(4\theta )\cdot \dfrac{d\theta }{dx}\; dx - \left(\int_{x\; :\; \theta = \frac{\pi }{3}}^{x\; :\; \theta = \frac{\pi }{4}}\pi \cdot \sin^2(4\theta )\cdot \dfrac{d\theta }{dx}\; dx\right) \\ &= \dfrac{41\pi }{120} - \dfrac{3\pi }{40} \\ &=\dfrac{4\pi }{15}. \end{aligned}$ Hence, the total volume of the solid of revolution is equal to $V_{\text{body}}+V_{\text{tail}}=\tfrac{12\pi }{5}+\tfrac{4\pi }{15}=\tfrac{8\pi }{3}$ , making the answer $8+3=\boxed{11}$ .

By substituting $r=\sqrt{x^2+y^2}$ and $\theta=\tan^{-1}{(\frac{y}{x})}$ ,

we have $\sqrt{x^2+y^2}=\frac{\sin(4\tan^{-1}{(\frac{y}{x})})}{\sin(\tan^{-1}{(\frac{y}{x})})}$

$=\cos(\tan^{-1}{(\frac{y}{x})})\cos(2\tan^{-1}{(\frac{y}{x})})$

$=-\frac{(\frac{y}{x}-1)(\frac{y}{x}+1)}{((\frac{y}{x})^2+1)^{\frac{3}{2}}}$

Squaring both sides,

$x^2 + y^2=-\frac{16 (x^3 - x y^2)^2}{(x^2 + y^2)^3}$

$(x^2 + y^2)^4 = 16 (x^3 - x y^2)^2$

$(x^2 + y^2)^2 = 4 (x^3 - x y^2)$ (Positive)

The equation can be rewritten as

$y=f(x)=\sqrt{-x^2+2\sqrt{x^2+2x^3}-2x}$ for the upper part above x-axis

Domain is $-\frac{1}{2}\leq x\leq 4$

$y=g(x)=\sqrt{-x^2-2\sqrt{x^2+2x^3}-2x}$ for the lower part above x-axis

Domain is $-\frac{1}{2}\leq x\leq 0$

So the volume is $\pi\int^4_{-\frac{1}{2}}(f(x))^2-\pi\int^0_{-\frac{1}{2}}(g(x))^2 = \frac{8}{3} \pi$

(As the outermost surd is removed due to it being squared, the integration is not difficult to be calculated)

The answer is $8+3=11$