If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximating error in calculating its volume.

Solution :

Let r be the radius of a sphere and \(\delta\)r be the error in measuring the radius. Then, r = 9 cm and \(\delta\)r = 0.03 cm.

Let V be the volume of the sphere. Then,

V = \({4\over 3}\pi r^3\)  \(\implies\) \(dV\over dr\) = \(4\pi r^2\)

\(\implies\) \(({dV\over dr})_{r = 9}\) = \(4\pi \times 9^2\) = \(324 \pi\)

Let \(\delta\)V be the error in V due to error in V due to error \(\delta\)r in r. Then,

\(\delta\)V = \(dV\over dr\) \(\delta\)r  \(\implies\)  \(\delta\)V =  \(324 \pi\times 0.03\) = \(9.72 \pi cm^3\)

---

Similar Questions

Find the approximate value of f(3.02), where f(x) = \(3x^2 + 5x + 3\).

Verify Rolle’s theorem for the function f(x) = \(x^2\) – 5x + 6 on the interval [2, 3].

It is given that for the function f(x) = \(x^3 – 6x^2 + ax + b\) on [1, 3], Rolles’s theorem holds with c = \(2 +{1\over \sqrt{3}}\). Find the values of a and b, if f(1) = f(3) = 0.

Find the point on the curve y = cos x – 1, x \(\in\) \([{\pi\over 2}, {3\pi\over 2}]\) at which tangent is parallel to the x-axis.