You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
volume of a cylinder
- Thread starter Ruiz007
- Start date
D
Deleted member 4993
Guest
Ruiz007 said:Hi I need help!! :x
I need step by step help with this problem... How may cubic yards of dirt must be dug to make a well 4 feet 6 inches in diameter and 42 feet deep? Thank you!
You should convert all the given dimensions to inches - then find the volume of the well.
What is the volume of the well in cubic inches? ..............................................(1)
How many cubic inches are there in a cubic yard?............................................(2)
Divide (1) by (2) to get your answer.
Not to appear contradictory, but I do not believe it is necessary to convert to inches (though, one certainly can if they wish). There are 27 cubic feet in a cubic yard, so just find the volume, as is, in cubic feet, then divide by 27.
Since the diameter is 4.5 feet, then the radius is 2.25 feet.
\(\displaystyle \frac{{\pi}(2.25)^{2}(42)}{27}\)
Since the diameter is 4.5 feet, then the radius is 2.25 feet.
\(\displaystyle \frac{{\pi}(2.25)^{2}(42)}{27}\)
D
Deleted member 4993
Guest
galactus said:Not to appear contradictory, but I do not believe it is necessary to convert to inches (though, one certainly can if they wish). There are 27 cubic feet in a cubic yard, so just find the volume, as is, in cubic feet, then divide by 27.
Since the diameter is 4.5 feet, then the radius is 2.25 feet.
\(\displaystyle \frac{{\pi}(2.25)^{2}(42)}{27}\)
Not at all contradictory - I just feel comfortable working with inches - I somehow visualize it better. As you have said it is just two different (correct) approaches to the same problem.
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
\(\displaystyle Without \ a \ calculator, \ which \ would \ you \ rather \ reduced:\)
\(\displaystyle \frac{42\pi(2.25)^2}{27}(ft.), \ \frac{504\pi(27)^2}{46656}(in.), \ or \ \frac{1280.16\pi(68.58)^2}{764554.857984}(cm.)?\)
\(\displaystyle The \ reason \ for \ my \ above \ statement \ is \ since \ Subhotosh \ Khan \ feels \ more \ comfortable \ working\)
\(\displaystyle with \ inches \ instead \ of \ feet; \ maybe \ he \ feels \ more \ comfortable \ still \ working \ with \ centimeters\)
\(\displaystyle instead \ of \ inches.\)
\(\displaystyle \frac{42\pi(2.25)^2}{27}(ft.), \ \frac{504\pi(27)^2}{46656}(in.), \ or \ \frac{1280.16\pi(68.58)^2}{764554.857984}(cm.)?\)
\(\displaystyle The \ reason \ for \ my \ above \ statement \ is \ since \ Subhotosh \ Khan \ feels \ more \ comfortable \ working\)
\(\displaystyle with \ inches \ instead \ of \ feet; \ maybe \ he \ feels \ more \ comfortable \ still \ working \ with \ centimeters\)
\(\displaystyle instead \ of \ inches.\)
Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.BigGlenntheHeavy said:\(\displaystyle Without \ a \ calculator, \ which \ would \ you \ rather \ reduce ?:\)
\(\displaystyle \frac{42\pi(2.25)^2}{27} \ or \ \frac{504\pi(27)^2}{46656}?\)
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
\(\displaystyle Denis, \ you're \ OK.\)