For this case we have that by definition, the volume of a cylinder is given by:
![V = \pi * r ^ 2 * h](https://img.qammunity.org/2020/formulas/mathematics/middle-school/9bbr1apmwdfg5ekvyuzuhy0h4dyqymtih6.png)
Where:
h: It is the height of the cylinder
A: It is the radius of the cylinder
According to the data we have:
![V = 9080 \ in ^ 3\\h = 20 \ in](https://img.qammunity.org/2020/formulas/mathematics/middle-school/kqz1bpj90i5y962cn6astego8g8356hili.png)
Substituting we have:
![9080 = \pi * r ^ 2 * 20](https://img.qammunity.org/2020/formulas/mathematics/middle-school/qwp84wft29i370io09p1387hds2rv3zem2.png)
Taking
![\pi = 3.14](https://img.qammunity.org/2020/formulas/mathematics/middle-school/3zfes56daztqslfd3xh6m403jrt9tjkkno.png)
![9080 = 62.80r ^ 2\\r ^ 2 = 144.59\\r = \pm \sqrt {144.59}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/jykm589co55pzfgz9mvpuo2dw2vfzk26vc.png)
We choose the positive value:
![r = \sqrt {144.59}\\r = 12.02](https://img.qammunity.org/2020/formulas/mathematics/middle-school/752im7a0rnpyy7orki5i59sg5789pus8b0.png)
Thus, the radius of the cylinder is
![r = 12.02 \ in](https://img.qammunity.org/2020/formulas/mathematics/middle-school/l1bbttixpzq1jmv50v0wj0b9wq0yghdefb.png)
Answer:
![12.02 \ in](https://img.qammunity.org/2020/formulas/mathematics/middle-school/6q14lczwtwnrcp1wx3d6nyyq2cwuj2uira.png)