Answer:
![2\pi r(r+rx)](https://img.qammunity.org/2021/formulas/physics/high-school/8mjnskir341rm8hdier1gnm9dmkvea54ux.png)
Step-by-step explanation:
Let the area of one circular side be given by the formula :
![A_(1) = \pi r^(2)](https://img.qammunity.org/2021/formulas/physics/high-school/mjlqpeuyekbd1ibvrq53g9rfz949nrpg06.png)
However, the wire is a solid cylinder, then it means that the total area is 2 ×
=
![2\pi r^(2)](https://img.qammunity.org/2021/formulas/physics/high-school/3ieq4pukb3l6hde1n5ztm4c6piin6zcvby.png)
However, there is the surface area to consider. This is the curved area of the wire. This is given as:
![A_(2) = lb](https://img.qammunity.org/2021/formulas/physics/high-school/h1zr7qp9ol90mdxdd52jdc9ctrnu0dktbi.png)
The length is x.
The breadth is calculated as follows - the length of the circle =
![\pi D = 2\pi r](https://img.qammunity.org/2021/formulas/physics/high-school/6i4q5lv91jzzyybol5k6ko251s53uyg1vj.png)
Then the area = lb
=
![2\pi rx](https://img.qammunity.org/2021/formulas/physics/high-school/33l8m8xod1mize0lwog22ry82dj683b88t.png)
Therefore, the total area is given as
![A_(1) + A_(2)](https://img.qammunity.org/2021/formulas/physics/high-school/2go2renl0hd26ezjgcx4tl5qd8p3zk2qac.png)
=
![2\pi r^(2) + 2\pi rx\\ 2\pi r(r+rx)](https://img.qammunity.org/2021/formulas/physics/high-school/zgwkn2ne97invy622tjqkqkqtbecn228vn.png)