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(A) Given radius $r = 7 	ext{ cm}$ and error in surface area $dA = 0.08 	ext{ cm}^2$. The surface area of a sphere is $A = 4 \pi r^2$. Differentiati

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$0.28$

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(A) Given radius $r = 7 	ext{ cm}$ and error in surface area $dA = 0.08 	ext{ cm}^2$. The surface area of a sphere is $A = 4 \pi r^2$. Differentiating with respect to $r$, we get $\frac{dA}{dr} = 8 \pi r$. Thus, $dA = 8 \pi r \cdot dr$. Substituting the values, $0.08 = 8 \pi (7) \cdot dr$. $dr = \frac{0.08}{56 \pi} = \frac{0.01}{7 \pi} 	ext{ cm}$. The volume of a sphere is $V = \frac{4}{3} \pi r^3$. Differentiating with respect to $r$, we get $\frac{dV}{dr} = 4 \pi r^2$. The approximate error in volume is $dV = \frac{dV}{dr} \cdot dr$. $dV = (4 \pi r^2) \cdot \left( \frac{0.01}{7 \pi} ight)$. Substituting $r = 7$, $dV = 4 \pi (7^2) \cdot \frac{0.01}{7 \pi} = 4 	imes 7 	imes 0.01 = 0.28 	ext{ cm}^3$.

The radius of a sphere is $7 \text{ cm}$. If an error of $0.08 \text{ cm}^2$ is made in measuring its surface area, then the approximate error (in $\text{cm}^3$) found in its volume is:

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