A circle is growing, its radius increasing by $4 \mathrm{~mm}$ per second. Find the rate at which the area is changing at the moment when the radius is $26 \mathrm{~mm}$ When the radius is $26 \mathrm{~mm}$, the area is changing at approximately (Round to the nearest thousandth as needed.)

Step 1 ：The area of a circle is given by the formula \(A = \pi r^2\).

Step 2 ：We are given that the radius is increasing at a rate of \(4 \mathrm{~mm/s}\), which we can denote as \(\frac{dr}{dt} = 4 \mathrm{~mm/s}\).

Step 3 ：We are asked to find the rate at which the area is changing, or \(\frac{dA}{dt}\), when the radius is \(26 \mathrm{~mm}\).

Step 4 ：We can find this by differentiating the area formula with respect to time \(t\), and then substituting the given values.

Step 5 ：By differentiating, we get \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\).

Step 6 ：Substituting the given values, we get \(\frac{dA}{dt} = 2\pi (26) (4)\).

Step 7 ：This simplifies to \(\frac{dA}{dt} = 208\pi \mathrm{~mm^2/s}\).

Step 8 ：Rounding to the nearest thousandth, we get \(\frac{dA}{dt} \approx 653.451 \mathrm{~mm^2/s}\).

Step 9 ：Final Answer: The rate at which the area is changing at the moment when the radius is \(26 \mathrm{~mm}\) is approximately \(\boxed{653.451 \mathrm{~mm^2/s}}\).