Answer:
or
![V=136.93\pi\ cm^3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/uyaj3z63tq4ybs5ug5gfdbf9gi8dps5xah.png)
Explanation:
The surface area of a sphere is:
![A_s=4\pi r^2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/dod9z6t1z388v4ftj2m6l17kuff6zpofwn.png)
Where r is the radius of the sphere
In this case we know that
![A_s = 275.561\ cm^2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ajgqoe857qj4p2kqez4hcrxgo7m4brnuxz.png)
So
![4\pi r^2=275.561](https://img.qammunity.org/2020/formulas/mathematics/middle-school/x2gl4vsl6bi1nf4110mrrd32horii6ffnh.png)
We solve the equation for r
![4\pi r^2=275.561\\r^2 = (275.561)/(4\pi)\\\\r=\sqrt{ (275.561)/(4\pi)}\\\\r=4.683\ cm](https://img.qammunity.org/2020/formulas/mathematics/middle-school/xrix4a3u82ye1a71ct91k20mqogrsimkmn.png)
Now we know the radius of the sphere.
The volume of a sphere is:
![V=(4)/(3)\pi r^3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/zn7p1lbtqa5xs0by5iezm20g8ifwtiean6.png)
We substitute the value of the radius in the formula
![V=(4)/(3)\pi (4.683)^3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/3gsgrcdcobvz8dj66efg8fyw5wieswup49.png)
![V=430.19\ cm^3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/lh0vu45fgecexyehmivq0e01es8fa9g0xy.png)