# If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating its surface area

**Solution:**

We can use differentials to calculate small changes in the dependent variable of a function corresponding to small changes in the independent variable

Let r be the radius of the sphere and Δr be the error in measuring the radius

Then, r = 9 m and Δr = 0.03 m

Now, the surface area S of the sphere is given by,

S = 4πr^{2}

Therefore,

On differentiating both sides wrt radius, we get

dS/dr = 8π r------(1)

Hence,

dS can also be written as:

dS = (dS/dr) Δr

On substituting the value of dS/dr from equation 1, we get

dS = (8π r) Δr

dS = 8π (9)(0.03)

dS = 2.16 π

Thus, the approximate error in calculating its surface area is 2.16π m^{2}

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.4 Question 7

## If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating its surface area.

**Summary:**

Given that the radius of a sphere is measured as 9 m with an error of 0.03 m. Hence, the approximate error in calculating its surface area is 2.16π m^{2}

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