Question

Simplify: pi * R^3 + ( - pi * (R-r)^3)

Answer 1

a^3-b^3=(a-b)(a^2+ab+b^2)
use it.

Answer 2

and take -pi as common?

Answer 3

yes

Answer 4

-pi ((r^3) - (R-r)^3) (R^2 + (R^2-Rr) + (R-r)^2)

Answer 5

omg how can this be the CENTER OF A MASS?

Answer 6

this is how i came to this position:

"Xcom = (m1*x1 + m2*x2)/ (m1 + m2)
Area of Big Circle (m1):
m1=Pi* R^2
x1= Center of mass: R
m1* x1 = pi * R^3
Therefore value of outer cirle has a full mass of m1 = pi * R^3
--------------------------
Area of Small Circle: (hole, therefore Negative) (m2)
m2= -pi * (R-r)^2
x2 = Center of mass: (R-r)
m2* x2 = - pi * (R-r)^3
I added up both Areas:
The small circle: A, and outer cirle with a small cirle cut out of it.: B,
I added up both: (m1*x1 + m2*x2)
Total (Mi*Xi) = pi * R^3 + ( - pi * (R-r)^3)"

---- mathematically did i do anything wrong?

Answer 7

\[\pi \times R ^{3} - \pi \times (R-r) ^{3} = \pi \times (R ^{3}-R^{3}+3R^{2}r+3Rr^{2}-r^{3})\]
\[\pi \times (3Rr(R+r)-r^{3})=\pi \times r(3R(R+r)-r^{2})=\pi \times 3R^{2}r+\pi \times 3Rr^{2}-\pi r^{3}\]

Answer 8

@immeen04 , I think Kathi26 solved it