AP Calculus BC 9.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve - FRQs - Exam Style Questions
Calc-Ok Question
Let \(S\) be the region bounded by the polar curve \(r(\theta)=3\sqrt{\theta}\,\sin(\theta^{2})\) for \(0\le \theta \le \sqrt{\pi}\), as shown.
(a) Find the area of \(S\).
(b) What is the average distance from the origin to a point on the curve \(r(\theta)=3\sqrt{\theta}\,\sin(\theta^{2})\) for \(0\le \theta \le \sqrt{\pi}\)?
(c) A line through the origin with positive slope \(m\) divides \(S\) into two regions of equal area. Write, but do not solve, an equation (involving one or more integrals) whose solution gives \(m\).
(d) For \(k>0\), let \(A(k)\) be the area of the portion of \(S\) that is also inside the circle \(r=k\cos\theta\). Find \(\displaystyle \lim_{k\to\infty} A(k)\).
Most-appropriate topic codes (CED):
• TOPIC 9.8: Area of a Polar Region (BC) — parts (a), (c), (d)
• TOPIC 9.9: Area Bounded by Two Polar Curves (BC) — part (d)
• TOPIC 8.1: Average Value of a Function — part (b)
• TOPIC 9.9: Area Bounded by Two Polar Curves (BC) — part (d)
• TOPIC 8.1: Average Value of a Function — part (b)
▶️ Answer/Explanation
(a) Area of \(S\)
Polar-area formula (single curve): \(\displaystyle \text{Area}=\frac12\int_{\alpha}^{\beta}\!\big(r(\theta)\big)^{2}\,d\theta\).
\[ A=\frac12\int_{0}^{\sqrt{\pi}}\!\!\big(3\sqrt{\theta}\,\sin(\theta^{2})\big)^{2} d\theta =\frac12\int_{0}^{\sqrt{\pi}}\! 9\theta\,\sin^{2}(\theta^{2})\,d\theta. \]
Use \(\sin^{2}u=\tfrac12(1-\cos 2u)\) with \(u=\theta^{2}\):
\[ A=\frac12\int_{0}^{\sqrt{\pi}}\!9\theta\cdot \frac{1-\cos(2\theta^{2})}{2}\,d\theta =\frac{9}{4}\int_{0}^{\sqrt{\pi}}\!\theta\,d\theta-\frac{9}{4}\cdot\frac12\int_{0}^{\sqrt{\pi}}\!\theta\cos(2\theta^{2})\,d\theta. \]
For \(\displaystyle \int \theta\cos(2\theta^{2})\,d\theta\), let \(w=\theta^{2}\Rightarrow dw=2\theta\,d\theta\):
\(\displaystyle \int \theta\cos(2\theta^{2})\,d\theta=\tfrac12\int \cos(2w)\,dw=\tfrac14\sin(2w)=\tfrac14\sin(2\theta^{2}).\)
Therefore an antiderivative for \(9\theta\sin^{2}(\theta^{2})\) is \(\displaystyle \frac{9}{4}\theta^{2}-\frac{9}{8}\sin(2\theta^{2})\). Hence
\[ A=\frac12\Big[\frac{9}{4}\theta^{2}-\frac{9}{8}\sin\!\big(2\theta^{2}\big)\Big]_{0}^{\sqrt{\pi}} =\frac12\cdot\frac{9}{4}\pi=\boxed{\frac{9\pi}{8}}\approx \mathbf{3.534292}. \]
(b) Average distance from the origin
Average value on \([0,\sqrt{\pi}]\): \(\displaystyle \text{Avg}=\frac{1}{\sqrt{\pi}}\int_{0}^{\sqrt{\pi}}\! r(\theta)\,d\theta\). (Topic 8.1.)
\[ \text{Avg}=\frac{1}{\sqrt{\pi}}\int_{0}^{\sqrt{\pi}}3\sqrt{\theta}\,\sin(\theta^{2})\,d\theta. \]
Substitute \(x=\theta^{2}\Rightarrow dx=2\theta\,d\theta\), \(d\theta=\frac{dx}{2\sqrt{x}}\), \(\sqrt{\theta}=x^{1/4}\):
\[ \text{Avg}=\frac{1}{\sqrt{\pi}}\int_{0}^{\pi}\frac{3}{2}\,x^{-1/4}\sin x\,dx, \] which has no elementary antiderivative; evaluating numerically (e.g., Simpson’s rule / calculator) gives
\[ \boxed{\text{Avg}\approx 1.579933\ \text{fish-units}} \ (\text{about } \mathbf{1.580}). \]
(c) Line through the origin dividing \(S\) into equal areas
A line with slope \(m>0\) through the origin corresponds to the polar angle \(\theta=\arctan m\).
Equal areas \(\Rightarrow\) area from \(0\) to \(\arctan m\) equals half of the total area:
\[ \boxed{\;\frac12\int_{0}^{\arctan m}\!\!\big(r(\theta)\big)^{2}\,d\theta \;=\;\frac12\cdot\frac12\int_{0}^{\sqrt{\pi}}\!\!\big(r(\theta)\big)^{2}\,d\theta\;} \] with \(r(\theta)=3\sqrt{\theta}\,\sin(\theta^{2})\).
(d) \(\displaystyle \lim_{k\to\infty}A(k)\) for the circle \(r=k\cos\theta\)
The circle \(r=k\cos\theta\) (center \((k/2,0)\), radius \(k/2\)) expands and, as \(k\to\infty\), includes all points with \(x\ge 0\), i.e., \(-\tfrac{\pi}{2}\le \theta\le \tfrac{\pi}{2}\).
Since \(S\) uses \(0\le \theta\le \sqrt{\pi}\) and \(\tfrac{\pi}{2}<\sqrt{\pi}\), the limiting overlap is the portion of \(S\) with \(0\le \theta\le \tfrac{\pi}{2}\).
Therefore \[ \lim_{k\to\infty}A(k) =\frac12\int_{0}^{\pi/2}\!\!\big(r(\theta)\big)^{2}d\theta =\frac12\int_{0}^{\pi/2}\! 9\theta\sin^{2}(\theta^{2})\,d\theta. \] Using the antiderivative from part (a), \[ \lim_{k\to\infty}A(k) =\frac12\Big[\frac{9}{4}\theta^{2}-\frac{9}{8}\sin(2\theta^{2})\Big]_{0}^{\pi/2} =\frac{9}{8}\left(\frac{\pi^{2}}{4}\right)-\frac{9}{16}\sin\!\Big(\frac{\pi^{2}}{2}\Big). \] Numerical value: \[ \boxed{\lim_{k\to\infty}A(k)\approx \mathbf{3.324}}. \]
Polar-area formula (single curve): \(\displaystyle \text{Area}=\frac12\int_{\alpha}^{\beta}\!\big(r(\theta)\big)^{2}\,d\theta\).
\[ A=\frac12\int_{0}^{\sqrt{\pi}}\!\!\big(3\sqrt{\theta}\,\sin(\theta^{2})\big)^{2} d\theta =\frac12\int_{0}^{\sqrt{\pi}}\! 9\theta\,\sin^{2}(\theta^{2})\,d\theta. \]
Use \(\sin^{2}u=\tfrac12(1-\cos 2u)\) with \(u=\theta^{2}\):
\[ A=\frac12\int_{0}^{\sqrt{\pi}}\!9\theta\cdot \frac{1-\cos(2\theta^{2})}{2}\,d\theta =\frac{9}{4}\int_{0}^{\sqrt{\pi}}\!\theta\,d\theta-\frac{9}{4}\cdot\frac12\int_{0}^{\sqrt{\pi}}\!\theta\cos(2\theta^{2})\,d\theta. \]
For \(\displaystyle \int \theta\cos(2\theta^{2})\,d\theta\), let \(w=\theta^{2}\Rightarrow dw=2\theta\,d\theta\):
\(\displaystyle \int \theta\cos(2\theta^{2})\,d\theta=\tfrac12\int \cos(2w)\,dw=\tfrac14\sin(2w)=\tfrac14\sin(2\theta^{2}).\)
Therefore an antiderivative for \(9\theta\sin^{2}(\theta^{2})\) is \(\displaystyle \frac{9}{4}\theta^{2}-\frac{9}{8}\sin(2\theta^{2})\). Hence
\[ A=\frac12\Big[\frac{9}{4}\theta^{2}-\frac{9}{8}\sin\!\big(2\theta^{2}\big)\Big]_{0}^{\sqrt{\pi}} =\frac12\cdot\frac{9}{4}\pi=\boxed{\frac{9\pi}{8}}\approx \mathbf{3.534292}. \]
(b) Average distance from the origin
Average value on \([0,\sqrt{\pi}]\): \(\displaystyle \text{Avg}=\frac{1}{\sqrt{\pi}}\int_{0}^{\sqrt{\pi}}\! r(\theta)\,d\theta\). (Topic 8.1.)
\[ \text{Avg}=\frac{1}{\sqrt{\pi}}\int_{0}^{\sqrt{\pi}}3\sqrt{\theta}\,\sin(\theta^{2})\,d\theta. \]
Substitute \(x=\theta^{2}\Rightarrow dx=2\theta\,d\theta\), \(d\theta=\frac{dx}{2\sqrt{x}}\), \(\sqrt{\theta}=x^{1/4}\):
\[ \text{Avg}=\frac{1}{\sqrt{\pi}}\int_{0}^{\pi}\frac{3}{2}\,x^{-1/4}\sin x\,dx, \] which has no elementary antiderivative; evaluating numerically (e.g., Simpson’s rule / calculator) gives
\[ \boxed{\text{Avg}\approx 1.579933\ \text{fish-units}} \ (\text{about } \mathbf{1.580}). \]
(c) Line through the origin dividing \(S\) into equal areas
A line with slope \(m>0\) through the origin corresponds to the polar angle \(\theta=\arctan m\).
Equal areas \(\Rightarrow\) area from \(0\) to \(\arctan m\) equals half of the total area:
\[ \boxed{\;\frac12\int_{0}^{\arctan m}\!\!\big(r(\theta)\big)^{2}\,d\theta \;=\;\frac12\cdot\frac12\int_{0}^{\sqrt{\pi}}\!\!\big(r(\theta)\big)^{2}\,d\theta\;} \] with \(r(\theta)=3\sqrt{\theta}\,\sin(\theta^{2})\).
(d) \(\displaystyle \lim_{k\to\infty}A(k)\) for the circle \(r=k\cos\theta\)
The circle \(r=k\cos\theta\) (center \((k/2,0)\), radius \(k/2\)) expands and, as \(k\to\infty\), includes all points with \(x\ge 0\), i.e., \(-\tfrac{\pi}{2}\le \theta\le \tfrac{\pi}{2}\).
Since \(S\) uses \(0\le \theta\le \sqrt{\pi}\) and \(\tfrac{\pi}{2}<\sqrt{\pi}\), the limiting overlap is the portion of \(S\) with \(0\le \theta\le \tfrac{\pi}{2}\).
Therefore \[ \lim_{k\to\infty}A(k) =\frac12\int_{0}^{\pi/2}\!\!\big(r(\theta)\big)^{2}d\theta =\frac12\int_{0}^{\pi/2}\! 9\theta\sin^{2}(\theta^{2})\,d\theta. \] Using the antiderivative from part (a), \[ \lim_{k\to\infty}A(k) =\frac12\Big[\frac{9}{4}\theta^{2}-\frac{9}{8}\sin(2\theta^{2})\Big]_{0}^{\pi/2} =\frac{9}{8}\left(\frac{\pi^{2}}{4}\right)-\frac{9}{16}\sin\!\Big(\frac{\pi^{2}}{2}\Big). \] Numerical value: \[ \boxed{\lim_{k\to\infty}A(k)\approx \mathbf{3.324}}. \]