How do you do this question?

V = 125π/6 h³

Step-by-step explanation:

First, let's draw a cross section.  If the radius of the sphere is R, and the radius of the cylindrical hole is r, and the height is 5a, then using Pythagorean theorem, we can write:

R² = r² + (5h/2)²

Now let's slice the sphere into concentric shells (like Russian nesting dolls).  Each shell is a hollow cylinder with a radius x, a thickness dx, and a height 2y.  The volume of each shell is:

dV = 2π r h t

dV = 2π (x) (2y) (dx)

dV = 4π x y dx

x² + y² = R², so:

dV = 4π x √(R² − x²) dx

The total volume is the sum of all the shells from x=r to x=R.

V = ∫ dV

V = ∫ 4π x √(R² − x²) dx

If u = R² − x², then du = -2x dx, and -2du = 4x dx.

V = ∫ -2π √u du

V = -2π ∫ √u du

V = -2π (⅔ u^(3/2)) + C

V = -4π/3 u^(3/2) + C

V = -4π/3 (R² − x²)^(3/2) + C

Evaluate from x=r to x=R.

V = -4π/3 (R² − R²)^(3/2) − -4π/3 (R² − r²)^(3/2)

V = 4π/3 (R² − r²)^(3/2)

Since R² = r² + (5h/2)², then R² − r² = (5h/2)².

V = 4π/3 ((5h/2)²)^(3/2)

V = 4π/3 (5h/2)³

V = 125π/6 h³

The volume of both napkin rings is the same, and does not depend on the radius of the sphere or the radius of the hole.

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents

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step-by-step explanation:

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