SOLUTION: A sphere has surface area of 1296 pi in^2. find the volume of the sphere. The choices are A)24 pi in^3 B)1944 pi in^3 C)432 pi in^3 D)7776 pi in^3

Algebra ->  Surface-area -> SOLUTION: A sphere has surface area of 1296 pi in^2. find the volume of the sphere. The choices are A)24 pi in^3 B)1944 pi in^3 C)432 pi in^3 D)7776 pi in^3      Log On


   

Question 218249: A sphere has surface area of 1296 pi in^2. find the volume of the sphere.
The choices are
A)24 pi in^3 B)1944 pi in^3
C)432 pi in^3 D)7776 pi in^3
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
A sphere has surface area of 1296 pi in^2. find the volume of the sphere.
The choices are

A)24 pi in^3 B)1944 pi in^3
C)432 pi in^3 D)7776 pi in^3

Step 1. The surface area S of a sphere is given as S=4%2Api%2Ar%5E2. So

S=1296=4%2Api%2Ar%5E2

Divide 4%2Api to both sides of the equation

1296%2F%284%2Api%29=r%5E2

Take the square root to both sides of the equation

sqrt%281296%2F%284%2Api%29%29=r

Step 2. The volume V of a sphere is given as V=4%2F3%2Api%2Ar%5E3. then

V=4%2F3%2Api%2Ar%5E3

V=4%2F3%2Api%2A%28sqrt%281296%2F%284%2Api%29%29%29%5E3

V=4%2F3%2Api%2A%281296%2F%284%2Api%29%29%5E%283%2F2%29 where sqrt%28x%29=x%5E%281%2F2%29 or %28sqrt%28x%29%29%5E3=x%5E%283%2F2%29

V=1%2F%283%2Asqrt%284%2Api%29%29%2A1296%5E%283%2F2%29 where I cancelled the 4%2Api in numerator leaving sqrt%284%2Api%29 in denominator

V=1296%5E%283%2F2%29%2Asqrt%28pi%29%2F%283%2A2%2Api%29 where I multiplied by sqrt%28pi%29%2Fsqrt%28pi%29=1 to get rid of square root of pi in denominator.

V=1296%5E%283%2F2%29%2Asqrt%28pi%29%2F%286%2Api%29

V=1296%5E%283%2F2%29%2Asqrt%28pi%29%2F%286%2Api%29

V=7776%2Asqrt%28pi%29%2Fpi

Step 3. ANSWER is V=7776%2Asqrt%28pi%29%2Fpi. I don't see this answer among your choices.

I hope the above steps were helpful.

For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

And good luck in your studies!

Respectfully,
Dr J

drjctu@gmail.com

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