Answer:
![1694 \pi](https://img.qammunity.org/2021/formulas/mathematics/middle-school/8xjnw7qsy9f3t7iam18lnlym0krr1wu17r.png)
Explanation:
I don't see a diagram, so I'm assuming that r is the radius and x is the height of a right circular cylinder (like a soup can).
The volume of the cylinder can be found from the formula
![V=\pi (\text{radius})^2 \cdot (\text{height})](https://img.qammunity.org/2021/formulas/mathematics/middle-school/1p2393nrda4xrm8thtat9tox0apznyfv99.png)
using the variables in your problem, this is
![V=\pi r^2x](https://img.qammunity.org/2021/formulas/mathematics/middle-school/mghxt2b3jk5z6jo4nw2br44ypzp6hdmt0b.png)
Plug in the given values for r and x.
![V=\pi(11)^2(14) \\ V=\pi(121)(14) \\ V=1694\pi](https://img.qammunity.org/2021/formulas/mathematics/middle-school/a8jxfuvag1ho4j55ihs56hf5ufdmxlnqo7.png)
If you need an approximate decimal value for the volume, use a value of pi (ask your teacher how accurate it should be).
Example: Using pi = 3.1416,
![V \approx (1694)(3.1416)=5321.8704](https://img.qammunity.org/2021/formulas/mathematics/middle-school/u1s3ms88p1hvaqyy3o6nv4kgja0dpoogsr.png)
You might be asked to round that answer, say to 5322 cubic units.