Question

sa \\o= 4\pi r^2 \quad sa\blacktriangle\\ = \pi r^2 + \pi rs
v \text{\ue845} \\ = \pi r^2 h \quad v\\o\\ \frac{4}{3}\pi r^3 \quad v\blacktriangle\\ = \frac{1}{3}\pi r^2 h
\underline{\hspace{10cm}}
\textbf{for problems 1-6:} write out the formula you will be using, show how you work out the problem for the values given and then write your answer with the appropriate label in the blank provided.
lt 2.6 i can find the surface area and volume of a cylinder and put the proper labels on my answers.
1.
\includegraphicsheight=5cm{cylinder1.png} \quad 3.14 \cdot
\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad sa = \underline{\hspace{2cm}}
\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad v = \underline{\hspace{2cm}}
2.
\includegraphicsheight=4cm{cylinder2.png}
\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad sa = \underline{\hspace{2cm}}
\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad v = \underline{\hspace{2cm}}

Explanation:

Step1: Recall Cylinder SA Formula

The surface area (SA) of a cylinder is given by \( SA = 2\pi r^2 + 2\pi rh \), where \( r \) is the radius and \( h \) is the height. For the first cylinder, \( r = 14 \, \text{mm} \), \( h = 40 \, \text{mm} \), and \( \pi \approx 3.14 \).

Step2: Calculate \( 2\pi r^2 \)

Substitute \( r = 14 \) and \( \pi \approx 3.14 \):
\( 2 \times 3.14 \times 14^2 = 2 \times 3.14 \times 196 = 1230.88 \, \text{mm}^2 \).

Step3: Calculate \( 2\pi rh \)

Substitute \( r = 14 \), \( h = 40 \), and \( \pi \approx 3.14 \):
\( 2 \times 3.14 \times 14 \times 40 = 3516.8 \, \text{mm}^2 \).

Step4: Sum for SA

Add the two parts: \( SA = 1230.88 + 3516.8 = 4747.68 \, \text{mm}^2 \).

Step5: Recall Cylinder Volume Formula

The volume (V) of a cylinder is \( V = \pi r^2 h \).

Step6: Calculate Volume

Substitute \( r = 14 \), \( h = 40 \), and \( \pi \approx 3.14 \):
\( V = 3.14 \times 14^2 \times 40 = 3.14 \times 196 \times 40 = 24617.6 \, \text{mm}^3 \).

(For problem 2, follow similar steps with \( r = \frac{4.5}{2} = 2.25 \, \text{yd} \), \( h = 19 \, \text{yd} \), \( \pi \approx 3.14 \):
SA: \( 2 \times 3.14 \times 2.25^2 + 2 \times 3.14 \times 2.25 \times 19 \approx 31.7925 + 268.47 = 300.2625 \, \text{yd}^2 \)
V: \( 3.14 \times 2.25^2 \times 19 \approx 3.14 \times 5.0625 \times 19 \approx 299.4375 \, \text{yd}^3 \))

Answer:

(for Problem 1):
SA = \( \boldsymbol{4747.68 \, \text{mm}^2} \)
V = \( \boldsymbol{24617.6 \, \text{mm}^3} \)

(For Problem 2, SA ≈ \( 300.26 \, \text{yd}^2 \), V ≈ \( 299.44 \, \text{yd}^3 \) (rounded appropriately))