This **article aims**Â to find the **area of a sector**. This **article uses the concept**Â of the **area of a sector**. The **reader should know how to find the area of the sector. Area of sector**Â of a circle is the amount of space enclosed within the boundary of the sector of circle. The **sector always starts from the center of the circle.**

The **area of sector**Â can be calculated using theÂ **following formulas:**

– **Area of a circular section**Â = $(\dfrac{\theta}{360^{\circ}}) \times \pi r ^ {2} $ where $ \theta $ is sector angle subtended by the arc at the **center in degrees**Â and $ r $ is the **radius of the circle**.

– **Area of a circular section**Â = $\dfrac {1} {2} \times r ^ {2} \theta $ where $ \theta $ is sector angle subtended by the arc at **center**Â and $ r $ is the **radius of the circle.**

**Expert Answer**

Let $ A $ represent the**Â area swept out**Â and $\theta $ the angle through which the **minute hand has turned.**

\[A = \dfrac {1} {2} r ^ {2} \theta \]

\[\dfrac { dA }{ dt } = \dfrac {1}{2} r ^ {2} \dfrac{ d\theta }{ dt }\]

We **know that:**

\[\dfrac {the\:area\: of \:sector }{the\: area\: of\: circle } = \dfrac { A }{ \pi r ^ {2} } \]

\[= \dfrac{ \theta }{2 \pi } Â \]

The **minute hand lasts**Â $ 60 $ **minutes per rotation**. Then the **angular velocity**Â is one**Â revolution per minute.**

\[\dfrac{d\theta }{dt} = \dfrac { 2\pi }{ 60 } = \dfrac { \pi }{ 30 } \dfrac { rad }{ min } \]

Thus

\[\dfrac{dA }{ dt } = \dfrac{1}{2} r^{2} \dfrac { d\theta }{ dt } = \dfrac { 1 }{ 2 } . (4)^{ 2 }. (\dfrac {\pi}{30}) \]

\[ = \dfrac{4\pi}{15} \dfrac{in^{2}}{min} \]

**Numerical Result**

**The area of sector that is swept out** is $ \dfrac{ 4\pi }{ 15 } \dfrac{ in ^ {2}}{min} $.

**Example**

**The minute hand of a particular clock is $ 5\: inches $ long. Starting when the hand points straight up, how fast does the area of the sector swept by the hand increase at each instant during the next hand revolution?**

**Solution**

**The $ A $ is given by:**

\[A = \dfrac{1} {2} r ^ {2} \theta \]

\[\dfrac { dA }{ dt } = \dfrac{ 1 }{ 2 } r ^ {2} \dfrac { d\theta}{ dt }\]

We **know that:**

\[\dfrac { the\:area\: of \:sector }{the\: area\: of\: circle } = \dfrac { A }{ \pi r ^ {2} } \]

\[= \dfrac{ \theta }{2 \pi } Â \]

The **minute hand lasts**Â $ 60 $ **minutes per rotation**. Then the **angular velocity**Â is one**Â revolution per minute.**

\[\dfrac{ d\theta }{ dt } = \dfrac{ 2\pi }{ 60 } = \dfrac{ \pi }{ 30 } \dfrac{ rad }{ min } \]

Thus

\[\dfrac{dA}{dt} = \dfrac{1}{2} r^{2} \dfrac{d\theta}{dt} = \dfrac{1}{2} . (5)^{2}. (\dfrac{\pi}{30}) \]

\[ = \dfrac{5\pi}{12} \dfrac{in^{2}}{min} \]

**The area of sector that is swept out**Â is $ \dfrac{5\pi}{12} \dfrac{in^{2}}{min} $.