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A spherical balloon is being inflated. Which of the following is an expression for △SA in terms of Δr ?

Hint: The surface area of a sphere is given by the equation SA=4πr².
a) ΔSA=4π(Δr)²
b) ΔSA=4πΔr
c) ΔSA=4π(r2​+r1​)Δr
d) ΔSA=4π(r2​−r1​)Δr

1 Answer

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The change in surface area of a spherical balloon (ΔSA) in terms of the change in radius (Δr) is best expressed as ΔSA=4π(r2​+r1​)Δr, where r1 and r2 are the initial and final radii of the sphere.

To answer the question about the change in surface area (ΔSA) of a spherical balloon in terms of the change in radius (Δr), we will need to use the formula for the surface area of a sphere, which is SA=4πr².

If we consider two radii, r1 and r2, with r2 being slightly larger due to inflation, the new surface area will be SA2=4πr2², and the original surface area is SA1=4πr1².

The change in surface area is ΔSA=SA2-SA1.

When we substitute the expressions for SA2 and SA1 into ΔSA and simplify, we find that:

ΔSA = 4π(r2² - r1²) = 4π(r2 + r1)(r2 - r1) = 4π(r2 + r1)Δr

Since r2 - r1 is the change in the radius (Δr), the expression for the change in surface area in terms of Δr is:

ΔSA = 4π(r2 + r1)Δr

Therefore, the correct option is c) ΔSA=4π(r2​+r1​)Δr.

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