Answer:
The surface area is equal to
![377\ in^(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/cmpt9d9fy854mzrqyhy8jitdf9sdlx7l29.png)
Explanation:
we know that
The surface area of the cylinder is equal to
![SA=2B+Ph](https://img.qammunity.org/2020/formulas/mathematics/middle-school/adutb8f2fvi7ftohtefzkqfzkxzq5w4gyy.png)
where
B is the area of the base of cylinder
P is the circumference of the base of cylinder
h is the height of the cylinder
Find the area of the base B
we have
![r=6\ in](https://img.qammunity.org/2020/formulas/mathematics/middle-school/g6cxvkr553gworzbktnrb2gv5nafjon6je.png)
substitute
![B=\pi (6^(2))=36\pi\ in^(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/9t2gvhtpjyhccdk7kgtino50vcegyi1yo7.png)
Find the circumference of the base
![P=2\pi r](https://img.qammunity.org/2020/formulas/mathematics/middle-school/f1f64ejgcwue597e7aub7n0y63sy9g9605.png)
we have
![r=6\ in](https://img.qammunity.org/2020/formulas/mathematics/middle-school/g6cxvkr553gworzbktnrb2gv5nafjon6je.png)
substitute
Find the surface area
we have
![B=36\pi\ in^(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/t896ns07o8bz69np83andk83eeqmv9ba76.png)
![h=4\ in](https://img.qammunity.org/2020/formulas/mathematics/middle-school/mmjhu792kg8lvtx16362z6a22gf0zbktcy.png)
substitute
![SA=2(36\pi)+(12\pi)(4)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/t37hdsu9c9nzo8b4i8s7x1kb1viv3giwjf.png)
![SA=72\pi+48\pi=120\pi=377\ in^(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/xh3mppphtceh50vi1ws1is1e7ntyflukmr.png)