Certainly! Let's break down the solution of the expression [tex]\(2 \pi r^2 + 2 \pi r h\)[/tex] when [tex]\(r = 3 \sqrt{2}\)[/tex] and [tex]\(h = 4 \sqrt{2}\)[/tex].
### Step-by-Step Solution:
1. Identify the given values:
- Radius ([tex]\(r\)[/tex]): [tex]\(3 \sqrt{2}\)[/tex]
- Height ([tex]\(h\)[/tex]): [tex]\(4 \sqrt{2}\)[/tex]
- [tex]\(\pi\)[/tex] (Pi): A constant approximately equal to 3.14159
2. Substitute [tex]\(r\)[/tex] and [tex]\(h\)[/tex] into the expression:
Substitute [tex]\(r = 3 \sqrt{2}\)[/tex] and [tex]\(h = 4 \sqrt{2}\)[/tex] into the expression [tex]\(2 \pi r^2 + 2 \pi r h\)[/tex].
3. Calculate [tex]\(r^2\)[/tex]:
[tex]\[ r^2 = (3 \sqrt{2})^2 = 9 \times 2 = 18 \][/tex]
4. Calculate [tex]\(2 \pi r^2\)[/tex]:
[tex]\[ 2 \pi r^2 = 2 \pi \times 18 = 36 \pi \][/tex]
5. Calculate [tex]\(2 \pi r h\)[/tex]:
Substitute the values of [tex]\(r\)[/tex] and [tex]\(h\)[/tex] into the expression:
[tex]\[ 2 \pi r h = 2 \pi \times (3 \sqrt{2}) \times (4 \sqrt{2}) \][/tex]
Simplify:
[tex]\[ 2 \pi r h = 2 \pi \times (3 \sqrt{2}) \times (4 \sqrt{2}) = 2 \pi \times 3 \times 4 \times 2 = 48 \pi \][/tex]
6. Sum the two terms to get the total:
[tex]\[
2 \pi r^2 + 2 \pi r h = 36 \pi + 48 \pi = 84 \pi
\][/tex]
### Numerical Evaluation:
7. Evaluate the numerical result:
Using the value of [tex]\(\pi\)[/tex], approximately 3.14159, evaluate [tex]\(84 \pi\)[/tex]:
[tex]\[
84 \pi \approx 84 \times 3.14159 = 263.8937829015427
\][/tex]
### Final Result:
Thus, the result of the expression [tex]\(2 \pi r^2 + 2 \pi r h\)[/tex] when [tex]\(r = 3 \sqrt{2}\)[/tex] and [tex]\(h = 4 \sqrt{2}\)[/tex] is approximately [tex]\(263.8937829015427\)[/tex].
Along the way, we had the intermediate calculations of:
- [tex]\(r = 3 \sqrt{2} \approx 4.242640687119286\)[/tex]
- [tex]\(h = 4 \sqrt{2} \approx 5.656854249492381\)[/tex]
- [tex]\(2 \pi r^2 \approx 113.09733552923258\)[/tex]
- [tex]\(2 \pi r h \approx 150.7964473723101\)[/tex]
Adding these intermediate results gives the total as:
[tex]\[ 263.8937829015427 \][/tex]
This detailed breakdown completes our solution.
