{
  "problem": "The polar curve $r(\\theta)=\\theta\\cdot\\sin(2\\theta)$ is graphed for $0\\leq\\theta\\leq\\pi$. Let $R$ be the region in the fourth quadrant enclosed by the curve and the $x$ -axis. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \\int^{\\scriptsize\\dfrac{\\pi}{2}}_{0}\\dfrac{1}{4}\\cdot\\theta^2\\cdot\\sin^2(2\\theta)\\,d\\theta$ (Choice B) B $ \\int_{\\scriptsize\\dfrac{\\pi}{2}}^{\\pi}\\dfrac{1}{4}\\cdot\\theta^2\\cdot\\sin^2(2\\theta)\\,d\\theta$ (Choice C) C $ \\int^{\\scriptsize\\dfrac{\\pi}{2}}_{0}\\dfrac{1}{2}\\cdot\\theta^2\\cdot\\sin^2(2\\theta)\\,d\\theta$ (Choice D) D $ \\int_{\\scriptsize\\dfrac{\\pi}{2}}^{\\pi}\\dfrac{1}{2}\\cdot\\theta^2\\cdot\\sin^2(2\\theta)\\,d\\theta$",
  "hints": [
    "This is the formula for the area enclosed by a polar curve $r(\\theta)$ between $\\theta=\\alpha$ and $\\theta=\\beta$ : $ \\int_{\\alpha}^{\\beta}\\dfrac{1}{2}\\left(r(\\theta)\\right)^{2}d\\theta$ We know $r(\\theta)$ but we still need to figure out $\\alpha$ and $\\beta$.",
    "Notice that the curve goes through the fourth quadrant, even though it's graphed for $0\\leq\\theta\\leq\\pi$. That's because for $\\theta$ -values between $\\dfrac{\\pi}{2}$ and $\\pi$, the sign of $r(\\theta)$ is negative. On that interval, the $\\theta$ -values for which $r(\\theta)=0$ are $0$, $\\dfrac{\\pi}{2}$, and $\\pi$. So $\\alpha$ must be $\\dfrac{\\pi}{2}$ and $\\beta$ must be $\\pi$.",
    "Let's plug ${r(\\theta)=\\theta\\cdot\\sin(2\\theta)}$, ${\\alpha=\\dfrac{\\pi}{2}}$, and ${\\beta=\\pi}$ into the formula and expand the parentheses: $\\begin{aligned} &\\phantom{=} \\int_{\\alpha}^{\\beta}\\dfrac{1}{2}\\left({r(\\theta)}\\right)^{2}d\\theta \\\\\\\\ &= \\int_{{\\scriptsize\\dfrac{\\pi}{2}}}^{{\\pi}}\\dfrac{1}{2}\\left({\\theta\\cdot\\sin(2\\theta)}\\right)^{2}d\\theta \\\\\\\\ &= \\int_{\\scriptsize\\dfrac{\\pi}{2}}^{\\pi}\\dfrac{1}{2}\\cdot\\theta^2\\cdot\\sin^2(2\\theta)\\,d\\theta \\end{aligned}$",
    "In conclusion, this integral represents the area of region $R$ : $ \\int_{\\scriptsize\\dfrac{\\pi}{2}}^{\\pi}\\dfrac{1}{2}\\cdot\\theta^2\\cdot\\sin^2(2\\theta)\\,d\\theta$"
  ]
}
{'source': 'AMPS', 'file': 'amps/khan/area-enclosed-by-polar-graphs/16075649591.json'}