Answer Posted / priya

But still no calculus is needed, well, not really...

Now, suppose you set out to answer the question in the absence of such specific direction as to how the answer should be given. We're going to set this sphere down on an x-axis centered right down the middle of the cylindrical hole.

let x=0 be the center of the sphere,
let r be the radius of the sphere, and
let h be the radius of the cylindrical hole.

The relationship between h and r, which we'll need later, is h² + 3² = r²

Now consider this slice of the holey sphere: Plane P is perpendicular to the axis of the hole at a directed distance of "x" from the center of the sphere, -3 ≤ x ≤ 3. The slice is the intersection of plane P with the sphere and its interior, minus the cylindrical hole: an annulus, the area bounded by a pair of concentric circles.

The radius of the smaller circle is h, and the radius of the larger circle is sqrt(r²-x²)

So the area of this slice is π(r²-x²)-π(h²) = π(r²-x² -h²)

Since r²=h²+3², (remember?), the area of this slice is π(h²+3²-x²-h²) = π(3²-x²)

This is the same as the area of a slice through a sphere of radius 3 with no hole, or with a hole whose radius is zero.

Since the cross-section area at distance x from the center of the holey sphere is the same as the cross-section area at the same distance from the center of an ordinary sphere of radius 3, it follows that the volume of the holey sphere, which is the integral of all these cross-section areas, is the same as the volume of the ordinary sphere.

The answer is (4/3)π(3³) = 36π

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1645

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4054

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1412

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1517

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1547

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1142

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1393

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1341

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1169

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1553

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1250

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1414

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1619

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1258

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1348

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