What Is the Area Bounded by $r=|\sin\theta|$?

The polar curve $r=|\sin\theta|$ has two lobes, one above and one below the origin. The area depends on whether you mean both enclosed components or just one: together they have area $\frac{\pi}{2}$, and each lobe has area $\frac{\pi}{4}$. The standard polar-area formula gives both results directly.

Use the polar-area formula

If a curve is described by $r=f(\theta)$ and traced once as $\theta$ runs from $a$ to $b$, the area it encloses is

$$A=\frac12\int_a^b r^2\,d\theta.$$

Here $r=|\sin\theta|$, so squaring removes the absolute value:

$$r^2=|\sin\theta|^2=\sin^2\theta.$$

The absolute value still matters for the shape: it keeps the radius nonnegative and makes the curve trace a lower lobe as well as an upper one. A full interval from $0$ to $2\pi$ traces both lobes once.

Calculate the total area

Substitute $r^2=\sin^2\theta$ into the formula and integrate over a full turn:

$$\begin{aligned} A&=\frac12\int_0^{2\pi}\sin^2\theta\,d\theta\\ &=\frac12\int_0^{2\pi}\frac{1-\cos(2\theta)}{2}\,d\theta\\ &=\frac14\left[\theta-\frac{\sin(2\theta)}{2}\right]_0^{2\pi}\\ &=\frac14(2\pi)=\frac{\pi}{2}. \end{aligned}$$

So the combined area of both lobes is $\frac{\pi}{2}$ square units.

Find the area of one lobe

For $0\leq\theta\leq\pi$, the curve traces the upper lobe. Its area is

$$A_{\text{upper}}=\frac12\int_0^\pi\sin^2\theta\,d\theta=\frac{\pi}{4}.$$

The interval $\pi\leq\theta\leq2\pi$ traces the lower lobe and gives the same area. Thus, if “the region” means a single connected component, the answer is $\frac{\pi}{4}$; if it means the entire area enclosed by the curve, the answer is $\frac{\pi}{2}$.

Check the shape in Cartesian coordinates

The equation also clarifies why there are two lobes. In polar coordinates, $y=r\sin\theta$. Since $r=|\sin\theta|$, we have $|y|=r^2$. Using $r^2=x^2+y^2$, this becomes

$$x^2+y^2=|y|.$$

For $y\geq0$, the equation is $x^2+(y-\frac12)^2=\frac14$, a circle centered at $(0,\frac12)$ with radius $\frac12$. For $y\leq0$, it is $x^2+(y+\frac12)^2=\frac14$, a circle centered at $(0,-\frac12)$ with the same radius. Each circle has area $\pi(\frac12)^2=\frac{\pi}{4}$, confirming the polar calculation.

Common mistakes

• Counting only one lobe without saying so. The upper component alone has area $\frac{\pi}{4}$, but the full curve has two components and total area $\frac{\pi}{2}$.
• Using the wrong angular interval. Integrating from $0$ to $\pi$ finds one lobe. Use $0$ to $2\pi$ for both.
• Forgetting to square the radius. The area formula uses $r^2$, not $r$. In this example, $|\sin\theta|^2=\sin^2\theta$.
• Adding the lobes twice. The interval $0$ to $2\pi$ traces each lobe once; no extra factor of two is needed.