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Start your free trial###### Bjørn Jakobsen

**Front End Web Development**Techdegree Graduate 17,003 Points

# Math.pow

// 4. Create a function that calculates the area of a circle. // The function should accept the radius of the circle as an argument // and return the area of that circle. // The area of a circle is the value of π * radius^2

My solution was

```
function areaCircle(radius) {
return Math.PI * radius * radius;
}
```

In the solution video Dave’s solution was

```
function areaCircle( radius ) {
return Math.PI * Math.pow(radius,2);
}
```

Both solutions works and I can’t figure out what Math.pow does why it’s better to use Dave’s solution. Is it just a way to calculate (radius * radius) without typing it twice or does it have any other advantages as well?

## 2 Answers

###### Steven Parker

228,991 PointsAs used here, the function just raises "radius" to the power of 2, which is mathematically equivalent to multiplying it by itself. While Dave's solution implements the formula more literally, yours is perfectly acceptable and will produce the same results.

###### Greg Radford

6,774 PointsJust to add to this, the reason you would use the Math.pow function over writing radius * radius is purely to simplify the code and allow easy use of variables. It would be difficult to manage:

```
radius * radius * radius * radius * radius * radius
```

vs

```
vs Math.pow(radius,6)
```

In this demo, the power is either 2 for the circle, or 3 for the sphere, so you could create a power variable, for example.

###### Steven Parker

228,991 PointsSince this function calculates an area, a power greater than 2 would never be required.

The sphere calculation requiring a power of 3 you are referring to is for __volume__, not area.