16. 극좌표계Polar Coordinates

Solution. (1) For $P(2,1)$, we have $r=\sqrt{x^2+y^2}=\sqrt{4+1}=\sqrt5$, and since the point lies in the first quadrant, $\theta=\tan^{-1}\dfrac12$. Hence the polar coordinates are $\left(\sqrt5,\ \tan^{-1}\dfrac12\right)$.

(2) For $P\left(2,\dfrac\pi3\right)$, $$x=r\cos\theta=2\cos\frac\pi3=2\cdot\frac12=1,\qquad y=r\sin\theta=2\sin\frac\pi3=2\cdot\frac{\sqrt3}2=\sqrt3.$$ Hence the Cartesian coordinates are $(1,\sqrt3)$.

Solution. (1) A representation with $r\gt0$ representing the same point as $\left(2,\dfrac\pi6\right)$ is obtained by adding an integer multiple of $2\pi$ to the angle, while a representation with $r\lt0$ is obtained by adding $\pi$ to the angle and then an integer multiple of $2\pi$. So for an integer $n$, $$\left(2,\ \frac\pi6+2n\pi\right),\qquad\left(-2,\ \frac{7\pi}6+2n\pi\right).$$

(2) Since $P(-2,0)$ has $x=-2$, $y=0$, we need $r\cos\theta=-2$, $r\sin\theta=0$. As $r\le0$ and $r\ne0$ (the point is not the origin), we need $\sin\theta=0$, so on $-\pi\le\theta\le2\pi$ we have $\theta\in\{-\pi,0,\pi,2\pi\}$. Solving $r\cos\theta=-2$ at each:
$\theta=0$: $\cos\theta=1\Rightarrow r=-2$ (satisfies $r\le0$)
$\theta=\pi$: $\cos\theta=-1\Rightarrow r=2$ (violates $r\le0$, discard)
$\theta=2\pi$: $\cos\theta=1\Rightarrow r=-2$ (satisfies $r\le0$)
$\theta=-\pi$: $\cos\theta=-1\Rightarrow r=2$ (violates $r\le0$, discard)
so the polar coordinates satisfying the condition are $$(-2,\ 0),\qquad(-2,\ 2\pi).$$

Solution. Substitute $x=r\cos\theta$, $y=r\sin\theta$ and simplify.

(1) $x=2 \Rightarrow r\cos\theta=2$.

(2) $xy=4 \Rightarrow (r\cos\theta)(r\sin\theta)=4 \Rightarrow r^2\cos\theta\sin\theta=4 \Rightarrow r^2\cdot\dfrac12\sin2\theta=4 \Rightarrow r^2\sin2\theta=8$.

(3) $x^2-y^2=1 \Rightarrow r^2\cos^2\theta-r^2\sin^2\theta=1 \Rightarrow r^2(\cos^2\theta-\sin^2\theta)=1 \Rightarrow r^2\cos2\theta=1$.

(4) Expanding $x^2+(y-3)^2=9$ gives $x^2+y^2-6y+9=9$, i.e., $x^2+y^2=6y$, so $r^2=6r\sin\theta$, that is, $r=6\sin\theta$ (the origin already satisfies this at $\theta=0$).

Solution. (1) $r\cos\theta=-4 \Rightarrow x=-4$.

(2) $r^2=4r\cos\theta \Rightarrow x^2+y^2=4x$ (multiply both sides by $r$ and simplify). Completing the square gives $(x-2)^2+y^2=4$, a circle of center $(2,0)$ and radius $2$.

(3) $r=1+2r\cos\theta \Rightarrow r=1+2x$, so squaring both sides gives $$x^2+y^2=(1+2x)^2=1+4x+4x^2 \ \Rightarrow\ y^2=3x^2+4x+1,$$ that is, $3x^2-y^2+4x+1=0$. (Completing the square gives $9\left(x+\dfrac23\right)^2-3y^2=1$, a hyperbola.)

(4) In $r=1-\cos\theta$, since $\cos\theta=x/r$, we get $r=1-\dfrac xr$, i.e., $r^2=r-x$. Since $r=\sqrt{x^2+y^2}$, this gives $x^2+y^2+x=\sqrt{x^2+y^2}$, and squaring both sides yields $$(x^2+y^2+x)^2=x^2+y^2.$$

(5) $\displaystyle r=\frac4{2\cos\theta-\sin\theta} \Rightarrow r(2\cos\theta-\sin\theta)=4 \Rightarrow 2r\cos\theta-r\sin\theta=4 \Rightarrow 2x-y=4$.

(6) In $r\sin\left(\dfrac{2\pi}3-\theta\right)=5$, $$\sin\left(\frac{2\pi}3-\theta\right)=\sin\frac{2\pi}3\cos\theta-\cos\frac{2\pi}3\sin\theta=\frac{\sqrt3}2\cos\theta+\frac12\sin\theta,$$ so $$r\left(\frac{\sqrt3}2\cos\theta+\frac12\sin\theta\right)=5 \Rightarrow \frac{\sqrt3}2r\cos\theta+\frac12r\sin\theta=5 \Rightarrow \frac{\sqrt3}2x+\frac12y=5,$$ that is, $\sqrt3x+y=10$.

Solution. (1) $r=1$: the unit circle of radius $1$ centered at the origin.

(2) $r=-1$: as $\theta$ ranges over all reals, the point is located at distance $|-1|=1$ from the origin in direction $\theta+\pi$, so this again traces the entire unit circle of radius $1$ centered at the origin — the same circle as in (1).

(3) $\theta=\pi/6$: since $r$ ranges over all reals, this is the entire line through the origin making angle $\pi/6$ with the positive $x$-axis (the part with $r\gt0$ lies in direction $\pi/6$, and the part with $r\lt0$ lies in the opposite direction $\pi/6+\pi=7\pi/6$).

(4) $\theta=7\pi/6$: likewise a line through the origin, and since $7\pi/6=\pi/6+\pi$, it represents exactly the same line as in (3).

Solution. (1) $1\le r\le2$, $0\le\theta\le\pi/2$: the annular sector between the circle of radius $1$ and the circle of radius $2$, both centered at the origin, restricted to the first quadrant — a sector of central angle $\pi/2$.

(2) $-3\le r\le2$, $\theta=\pi/4$: a segment on the line $y=x$. The part with $r\in[0,2]$ lies in direction $\pi/4$ (first quadrant) out to distance $2$, and the part with $r\in[-3,0]$ lies in the opposite direction $5\pi/4$ (third quadrant) out to distance $3$, so together this is the segment from $\left(-\dfrac{3\sqrt2}2,-\dfrac{3\sqrt2}2\right)$ to $(\sqrt2,\sqrt2)$.

(3) $2\pi/3\le\theta\le5\pi/6$: the unbounded sector (over all $r\ge0$) between angle $2\pi/3$ ($120°$) and angle $5\pi/6$ ($150°$), lying in the second quadrant.

Solution. (1) Since $\cos(-\theta)=\cos\theta$, the equation is unchanged when $\theta$ is replaced by $-\theta$, so $r=1-\cos\theta$ is symmetric about the $x$-axis. At $\theta=0$, $r=0$, so the curve has a cusp at the origin. As $\theta$ increases from $0$ to $\pi$, $r$ increases from $0$ to $2$, tracing the upper half: $\theta=\pi/2$ gives $r=1$ (point $(0,1)$), and $\theta=\pi$ gives $r=2$ (point $(-2,0)$). By the $x$-axis symmetry, the lower half is the reflection of this. Overall this is a **cardioid**, with a cusp at the origin, extending out to $(-2,0)$.

(2) Replacing $\theta$ by $-\theta$ leaves the equation unchanged since $\cos(-\theta)=\cos\theta$, so the curve is symmetric about the $x$-axis. Replacing $r$ by $-r$ also leaves it unchanged since $(-r)^2=r^2$, so the curve is symmetric about the origin (the pole) as well. Having both $x$-axis symmetry and symmetry about the origin, it is also symmetric about the $y$-axis. Since $\cos\theta\ge0$ is required, the curve (using the direct representation) is defined on $-\pi/2\le\theta\le\pi/2$; at $\theta=0$, $r=\pm2$, giving the points $(2,0)$ and $(-2,0)$, and at $\theta=\pm\pi/2$, $r=0$. The curve consists of two symmetric lobes meeting at the origin, extending along the $x$-axis toward $(2,0)$ and $(-2,0)$ respectively.

Solution. (1) $r^2=\sin2\theta$: replacing $r$ by $-r$ leaves the equation unchanged, so the curve is symmetric about the origin. Replacing $\theta$ by $\pi/2-\theta$ gives $\sin(2(\pi/2-\theta))=\sin(\pi-2\theta)=\sin2\theta$, unchanged, so the curve is symmetric about the line $\theta=\pi/4$ ($y=x$) (combined with the origin symmetry, it is also symmetric about the direction $\theta=5\pi/4$). Since $\sin2\theta\ge0$ is required, the curve is defined for $\theta\in[0,\pi/2]\cup[\pi,3\pi/2]$, with maximum $r=\pm1$ at $\theta=\pi/4$. This is a two-lobed **lemniscate** extending along the direction $y=x$ and its opposite direction.

(2) $r=\sin2\theta$: a **rose curve** with $n=2$; since $n$ is even, it has $2n=4$ petals. On $\theta\in[0,\pi/2]$, $r\ge0$ with maximum $r=1$ at $\theta=\pi/4$, giving a petal of length $1$ in the direction $\theta=\pi/4$ (the bisector of the first quadrant); likewise there are petals of length $1$ in the directions $\theta=3\pi/4,5\pi/4,7\pi/4$, forming a **four-petaled rose** along the four diagonal directions.

(3) $\displaystyle r=\frac12+\cos\theta$: since the coefficient of $\cos\theta$ ($1$) exceeds the constant term ($1/2$), this is a **limaçon with an inner loop**. Since $\cos(-\theta)=\cos\theta$, it is symmetric about the $x$-axis. $r=0$ occurs when $\cos\theta=-1/2$, i.e., $\theta=2\pi/3,4\pi/3$. On $\theta\in(-2\pi/3,2\pi/3)$, $r\ge0$, tracing a large outer loop with maximum $r=3/2$ at $\theta=0$; on $\theta\in(2\pi/3,4\pi/3)$, $r\lt0$ (e.g. $r=-1/2$ at $\theta=\pi$), tracing a small inner loop that passes through the origin and bulges out along the positive $x$-direction.

(4) $r=1-2\sin3\theta$: replacing $\theta$ by $\pi-\theta$ gives $\sin(3(\pi-\theta))=\sin(3\pi-3\theta)=\sin3\theta$, unchanged, so the curve is symmetric about the $y$-axis. Since $\sin3\theta\in[-1,1]$, we have $r\in[-1,3]$. At $\theta=\pi/2$, $\sin3\theta=\sin(3\pi/2)=-1$, so $r=3$ is maximal, corresponding to the point $(0,3)$ on the $y$-axis. At $\theta=\pi/6$, $\sin3\theta=1$, so $r=-1$ is minimal. $r=0$ occurs when $\sin3\theta=1/2$, i.e., $3\theta=\pi/6+2k\pi$ or $5\pi/6+2k\pi$. Where $r$ is negative, small loops through the origin appear on the opposite side, so this is a limaçon-type curve with a large outer loop symmetric about the $y$-axis interspersed with small inner loops through the origin.

(5) $r=\theta$: an **Archimedean spiral**. For $\theta\ge0$, $r$ increases at the same rate as $\theta$, so the curve starts at the origin, spirals counterclockwise, and moves steadily outward, with the radius increasing by a constant $2\pi$ each full revolution ($\Delta\theta=2\pi$), so consecutive coils are equally spaced. Allowing $\theta\lt0$ gives $r=\theta\lt0$, producing a second spiral symmetric in the opposite direction (angle $\theta+\pi$).

(6) $r=e^{\theta/10}$: a **logarithmic (equiangular) spiral**. We have $r\gt0$ for all $\theta$; as $\theta\to\infty$, $r\to\infty$, and as $\theta\to-\infty$, $r\to0^+$, so the curve approaches the origin arbitrarily closely without ever reaching it. Each time $\theta$ increases by $2\pi$, $r$ scales by a factor of $e^{\pi/5}$, so the spiral winds counterclockwise while expanding geometrically.

Solution. (1) $r=2\sec\theta$ means $r\cos\theta=2$, i.e., the line $x=2$. For fixed $\theta\in(-\pi/2,\pi/2)$, as $r$ ranges from $0$ to $2\sec\theta$, the $x$-coordinate $x=r\cos\theta$ of the corresponding point ranges from $0$ to $2$. As $\theta$ approaches $-\pi/2$ or $\pi/2$, $y=r\tan\theta$ diverges to $-\infty$ or $+\infty$ respectively, so as $\theta$ ranges over the whole interval, $y$ takes every real value. Hence this region is the entire vertical strip $\{(x,y):0\le x\le2\}$ between the $y$-axis and the line $x=2$ (with the boundary $x=0$ containing only the origin).

(2) Since $\cos\theta\ge0$ is required, $\theta\in[-\pi/2,\pi/2]$, and on this range $|r|\le\sqrt{\cos\theta}$. This is the filled-in interior of the curve $r^2=\cos\theta$, which has the same shape as the curve $r^2=4\cos\theta$ from Example 03(2) but with maximum radius $1$. That is, it consists of two symmetric filled lobes meeting at the origin, extending along the $x$-axis toward $(1,0)$ and $(-1,0)$ respectively.

Solution. First check whether the curves meet at the origin (the pole): $r=2\cos(\theta/3)=0$ holds at $\theta=3\pi/2$ (and its periodic values), but $r=\sqrt2$ is never $0$, so the two curves do not meet at the origin.

At an intersection point the distances from the origin must agree, so $|2\cos(\theta/3)|=\sqrt2$, i.e., $\cos^2(\theta/3)=\dfrac12$, $\cos(\theta/3)=\pm\dfrac{\sqrt2}2$.

Now, when $\theta$ increases by $3\pi$, the sign of $r=2\cos(\theta/3)$ flips and the angle rotates by $3\pi\equiv\pi\ (\mathrm{mod}\ 2\pi)$, so the same point is traced. Hence it suffices to consider an interval of length $3\pi$, say $\theta\in\left(-\dfrac{3\pi}2,\dfrac{3\pi}2\right]$.

If $\cos(\theta/3)=\dfrac{\sqrt2}2$, then $\dfrac\theta3=\pm\dfrac\pi4+2n\pi$, i.e., $\theta=\pm\dfrac{3\pi}4+6n\pi$, giving $\theta=\dfrac{3\pi}4,-\dfrac{3\pi}4$ within this interval. If $\cos(\theta/3)=-\dfrac{\sqrt2}2$, then $\dfrac\theta3=\pm\dfrac{3\pi}4+2n\pi$, i.e., $\theta=\pm\dfrac{9\pi}4+6n\pi$; within the interval this gives $\theta=\dfrac{9\pi}4-3\pi=-\dfrac{3\pi}4$ and $\theta=-\dfrac{9\pi}4+3\pi=\dfrac{3\pi}4$, where $r=2\cos(\theta/3)=-\sqrt2\lt0$; checking the actual location (distance $\sqrt2$, angle $\theta+\pi$) shows these coincide exactly with the two points already found.

Hence there are exactly two distinct intersection points. At $\theta=3\pi/4$, $r=2\cos(\pi/4)=\sqrt2$, and at $\theta=-3\pi/4$, $r=2\cos(-\pi/4)=\sqrt2$, so the intersection points are $$\left(\sqrt2,\ \frac{3\pi}4\right),\qquad\left(\sqrt2,\ -\frac{3\pi}4\right).$$ (In Cartesian coordinates, $(-1,1)$ and $(-1,-1)$ respectively.)

Solution. We use the tangent-slope formula for polar curves, $$\frac{dy}{dx}=\frac{\dfrac{dr}{d\theta}\sin\theta+r\cos\theta}{\dfrac{dr}{d\theta}\cos\theta-r\sin\theta}.$$

(1) For $r=-1+\cos\theta$, $\dfrac{dr}{d\theta}=-\sin\theta$. At $\theta=\pi/2$, $r=-1+\cos\dfrac\pi2=-1$ and $\dfrac{dr}{d\theta}=-\sin\dfrac\pi2=-1$, so $$\frac{dy}{dx}=\frac{(-1)\sin\frac\pi2+(-1)\cos\frac\pi2}{(-1)\cos\frac\pi2-(-1)\sin\frac\pi2}=\frac{-1+0}{0+1}=-1.$$ So the slope of the tangent line is $-1$.

(2) For $r=\cos2\theta$, $\dfrac{dr}{d\theta}=-2\sin2\theta$. At $\theta=0$, $r=\cos0=1$ and $\dfrac{dr}{d\theta}=-2\sin0=0$, so $$\frac{dy}{dx}=\frac{0\cdot\sin0+1\cdot\cos0}{0\cdot\cos0-1\cdot\sin0}=\frac10.$$ Since the denominator is $0$ while the numerator is not, the tangent line is vertical. This point is $(r,\theta)=(1,0)$, i.e., $(1,0)$ in Cartesian coordinates, so the tangent line is $x=1$.

Solution. For $r=\sin\theta$, $$x=r\cos\theta=\sin\theta\cos\theta=\frac12\sin2\theta,\qquad y=r\sin\theta=\sin^2\theta=\frac{1-\cos2\theta}2,$$ so $$\frac{dx}{d\theta}=\cos2\theta,\qquad\frac{dy}{d\theta}=\sin2\theta,$$ and hence $$\frac{dy}{dx}=\frac{\sin2\theta}{\cos2\theta}=\tan2\theta.$$ Differentiating this with respect to $\theta$ gives $\dfrac d{d\theta}\left(\dfrac{dy}{dx}\right)=2\sec^22\theta$, so $$\frac{d^2y}{dx^2}=\frac{\dfrac d{d\theta}\left(\dfrac{dy}{dx}\right)}{dx/d\theta}=\frac{2\sec^22\theta}{\cos2\theta}=\frac2{\cos^32\theta}.$$ At $\theta=\pi/6$, $\cos\dfrac{2\pi}6=\cos\dfrac\pi3=\dfrac12$, so $$\left.\frac{d^2y}{dx^2}\right|_{\theta=\pi/6}=\frac2{(1/2)^3}=\frac2{1/8}=16\gt0.$$ Since $d^2y/dx^2\gt0$, the polar curve $r=\sin\theta$ is **concave up** at $\theta=\pi/6$.

Solution. We use the arc length formula for polar curves, $L=\displaystyle\int\sqrt{r^2+\left(\dfrac{dr}{d\theta}\right)^2}\,d\theta$. The cardioid $r=1-\cos\theta$ is traced once completely on $\theta\in[0,2\pi]$, and $\dfrac{dr}{d\theta}=\sin\theta$, so $$r^2+\left(\frac{dr}{d\theta}\right)^2=(1-\cos\theta)^2+\sin^2\theta=1-2\cos\theta+\cos^2\theta+\sin^2\theta=2-2\cos\theta=4\sin^2\frac\theta2$$ (using $1-\cos\theta=2\sin^2(\theta/2)$). Since $\sin(\theta/2)\ge0$ on $\theta\in[0,2\pi]$, we have $\sqrt{4\sin^2(\theta/2)}=2\sin(\theta/2)$, so $$L=\int_0^{2\pi}2\sin\frac\theta2\,d\theta=\left[-4\cos\frac\theta2\right]_0^{2\pi}=-4\cos\pi-(-4\cos0)=4+4=8.$$

Solution. We use the polar area formula $A=\dfrac12\displaystyle\int r^2\,d\theta$.

(1) $r=2(1+\cos\theta)$ is traced once on $\theta\in[0,2\pi]$, so $$A=\frac12\int_0^{2\pi}4(1+\cos\theta)^2\,d\theta=2\int_0^{2\pi}(1+2\cos\theta+\cos^2\theta)\,d\theta.$$ Since $\displaystyle\int_0^{2\pi}1\,d\theta=2\pi$, $\displaystyle\int_0^{2\pi}2\cos\theta\,d\theta=0$, and $\displaystyle\int_0^{2\pi}\cos^2\theta\,d\theta=\pi$, $$A=2(2\pi+0+\pi)=2\cdot3\pi=6\pi.$$

(2) The region enclosed by $r=\theta$ $(0\le\theta\le\pi)$ and the $x$-axis corresponds to $\theta\in[0,\pi]$, so $$A=\frac12\int_0^\pi\theta^2\,d\theta=\frac12\left[\frac{\theta^3}3\right]_0^\pi=\frac12\cdot\frac{\pi^3}3=\frac{\pi^3}6.$$

Solution. The circle $r=1$ and the cardioid $r=1-\cos\theta$ meet where $1=1-\cos\theta$, i.e., $\cos\theta=0$, so $\theta=\pm\pi/2$. On $\theta\in\left(-\dfrac\pi2,\dfrac\pi2\right)$, $1-\cos\theta\lt1$, so the circle lies outside the cardioid, while on $\theta\in\left(\dfrac\pi2,\dfrac{3\pi}2\right)$, $1-\cos\theta\gt1$, so the cardioid lies outside the circle, leaving no region that is "inside the circle and outside the cardioid." Hence the desired region occurs on $\theta\in\left[-\dfrac\pi2,\dfrac\pi2\right]$, where it lies inside the circle $r=1$ and outside the cardioid $r=1-\cos\theta$, with area $$A=\frac12\int_{-\pi/2}^{\pi/2}\Big[1^2-(1-\cos\theta)^2\Big]\,d\theta.$$ Since $1-(1-\cos\theta)^2=1-(1-2\cos\theta+\cos^2\theta)=2\cos\theta-\cos^2\theta$, $$A=\frac12\int_{-\pi/2}^{\pi/2}(2\cos\theta-\cos^2\theta)\,d\theta=\int_{-\pi/2}^{\pi/2}\cos\theta\,d\theta-\frac12\int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta.$$ Since $\displaystyle\int_{-\pi/2}^{\pi/2}\cos\theta\,d\theta=\big[\sin\theta\big]_{-\pi/2}^{\pi/2}=1-(-1)=2$ and $\displaystyle\int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta=\left[\frac\theta2+\frac{\sin2\theta}4\right]_{-\pi/2}^{\pi/2}=\frac\pi4-\left(-\frac\pi4\right)=\frac\pi2$, $$A=2-\frac12\cdot\frac\pi2=2-\frac\pi4.$$

Solution. (1) For $r=\tan\theta$, $x=r\cos\theta=\sin\theta$ and $y=r\sin\theta=\dfrac{\sin^2\theta}{\cos\theta}$. As $\theta\to\dfrac\pi2^-$, $x=\sin\theta\to1$ while $\cos\theta\to0^+$ and $\sin^2\theta\to1$, so $y\to+\infty$, giving the vertical asymptote $x=1$. Since $\tan\theta$ is odd, $x(-\theta)=-x(\theta)$ and $y(-\theta)=y(\theta)$, so the curve is symmetric about the $y$-axis, and hence as $\theta\to-\dfrac\pi2^+$, $x\to-1$ and $y\to+\infty$, so $x=-1$ is also an asymptote. At $\theta=0$ the curve passes through the origin, and since $y=\sin^2\theta/\cos\theta\ge0$, the entire curve lies on or above the $x$-axis. The curve starts at the origin and splits into two symmetric branches rising toward the asymptotes $x=1$ and $x=-1$. The asymptotes are $x=1$ and $x=-1$.

(2) The point $(1,\pi/4)$ has $r=\tan\dfrac\pi4=1$ at $\theta=\pi/4$, so it is indeed on the curve, with Cartesian coordinates $x=\cos\dfrac\pi4=\dfrac{\sqrt2}2$, $y=\sin\dfrac\pi4=\dfrac{\sqrt2}2$. The line through this point parallel to the $x$-axis is $y=\dfrac{\sqrt2}2$.

By (1) the curve is symmetric about the $y$-axis, and since $\dfrac{dy}{d\theta}=\dfrac{\sin\theta(1+\cos^2\theta)}{\cos^2\theta}\gt0$ for $0\lt\theta\lt\pi/2$, $y=\sin^2\theta/\cos\theta$ is strictly increasing on $\theta\in[0,\pi/2)$. So on $\theta\in[0,\pi/4]$ the curve runs from the origin $(0,0)$ to $\left(\dfrac{\sqrt2}2,\dfrac{\sqrt2}2\right)$, and the symmetric branch on $\theta\in[-\pi/4,0]$ runs from $\left(-\dfrac{\sqrt2}2,\dfrac{\sqrt2}2\right)$ to the origin. The line $y=\dfrac{\sqrt2}2$ joins the endpoints of these two branches to form the upper boundary, giving a region with a cusp at the origin and closed above by the segment $y=\dfrac{\sqrt2}2$ $\left(-\dfrac{\sqrt2}2\le x\le\dfrac{\sqrt2}2\right)$.

By symmetry, we compute the area of the right half and double it. Since $x=\sin\theta$, $dx=\cos\theta\,d\theta$, so $$\int_0^{\sqrt2/2}y\,dx=\int_0^{\pi/4}\frac{\sin^2\theta}{\cos\theta}\cos\theta\,d\theta=\int_0^{\pi/4}\sin^2\theta\,d\theta=\left[\frac\theta2-\frac{\sin2\theta}4\right]_0^{\pi/4}=\frac\pi8-\frac14.$$ The rectangle under the line for $x\in\left[0,\dfrac{\sqrt2}2\right]$ has area $\dfrac{\sqrt2}2\cdot\dfrac{\sqrt2}2=\dfrac12$, so the area between the curve and the line on the right half is $$\frac12-\left(\frac\pi8-\frac14\right)=\frac34-\frac\pi8.$$ By symmetry, the total area is $$A=2\left(\frac34-\frac\pi8\right)=\frac32-\frac\pi4.$$

Solution. The average value of $r$ with respect to $\theta$ is $\operatorname{av}(r)=\dfrac1{\beta-\alpha}\displaystyle\int_\alpha^\beta r\,d\theta$. The cardioid $r=a(1-\cos\theta)$ is traced once completely on $\theta\in[0,2\pi]$, so $$\operatorname{av}(r)=\frac1{2\pi}\int_0^{2\pi}a(1-\cos\theta)\,d\theta=\frac a{2\pi}\Big[\theta-\sin\theta\Big]_0^{2\pi}=\frac a{2\pi}\big[(2\pi-0)-(0-0)\big]=\frac a{2\pi}\cdot2\pi=a.$$