An open hemispherical storage tank has radius 13 m. Oil flows into the tank such that the depth '$h$' of oil in the tank changes at the rate of 3 m / hr. When the depth $h = 1$m, the rate of change of the area of the top surface of the oil is

Let $R$ be the radius of the hemispherical tank, so $R = 13$ m.

Let $r$ be the radius of the top surface of the oil at depth $h$. The distance from the center of the hemisphere to the oil surface is $R - h$.

Using the Pythagorean theorem in the cross-section of the hemisphere:

$r^2 + (R - h)^2 = R^2$

$r^2 = R^2 - (R^2 - 2Rh + h^2) = 2Rh - h^2$

The area of the top surface of the oil is $A = \pi r^2$.

$A = \pi(2Rh - h^2)$

Differentiating with respect to time $t$:

$\dfrac{dA}{dt} = \pi(2R - 2h)\dfrac{dh}{dt}$

Given $R = 13$ m, $h = 1$ m, and $\dfrac{dh}{dt} = 3$ m/hr, substituting these values gives:

$\dfrac{dA}{dt} = \pi(2(13) - 2(1))(3)$

$\dfrac{dA}{dt} = \pi(26 - 2)(3)$

$\dfrac{dA}{dt} = 72\pi$ m$^2$/hr

Answer: $72\pi$ m$^2$ / hr