Calculate the surface area of water in contact with a semi-cylindrical trough of radius 63 cm and length 3 m when it is full.

The radius is given as $63 \text{ cm}$. The length is given as $3 \text{ m}$.
Convert the length to centimeters:

$$L = 3 \text{ m} = 3 \times 100 \text{ cm} = 300 \text{ cm}$$

So, $r = 63 \text{ cm}$ and $L = 300 \text{ cm}$.

When the semi-cylindrical trough is full, the water is in contact with two parts of the trough:

• The curved bottom surface of the semi-cylinder.
• The two semi-circular end surfaces.

• The curved surface area of a semi-cylinder is half the curved surface area of a full cylinder:

$$A_{\text{curved}} = \frac{1}{2} (2\pi r L) = \pi r L$$

• The area of the two semi-circular ends is equivalent to the area of one full circle:

$$A_{\text{ends}} = 2 \times \left(\frac{1}{2} \pi r^2\right) = \pi r^2$$

The total surface area ($A$) is the sum of the curved surface area and the area of the two semi-circular ends:

$$A = A_{\text{curved}} + A_{\text{ends}} = \pi r L + \pi r^2$$

Factor out $\pi r$:

$$A = \pi r (L + r)$$

Substitute the values $r = 63 \text{ cm}$ and $L = 300 \text{ cm}$:

$$A = \pi (63 \text{ cm}) (300 \text{ cm} + 63 \text{ cm})$$ $$A = \pi (63) (363) \text{ cm}^2$$ $$A = 22869 \pi \text{ cm}^2$$

Using the approximation $\pi \approx 3.14159$:

$$A \approx 22869 \times 3.14159$$ $$A \approx 71848.05 \text{ cm}^2$$

The surface area of the trough in contact with water when it is full is $\boxed{\text{71848.05 cm}^2}$.