Consider a parameterized curve $ C: x \equal{} e^{ \minus{} t}\cos t,\ y \equal{} e^{ \minus{} t}\sin t\ \left(0\leq t\leq \frac {\pi}{2}\right).$

(1) Find the length $ L$ of $ C$.

(2) Find the area $ S$ of the region bounded by $ C$, the $ x$ axis and $ y$ axis.

You may not use the formula $ \boxed{\int_a^b \frac {1}{2}r(\theta)^2d\theta }$ here.
### Part (1): Finding the Length \( L \) of \( C \)

1. **Compute the derivatives:**
   \[
   \frac{dx}{dt} = -e^{-t}\cos t - e^{-t}\sin t = -e^{-t}(\cos t + \sin t)
   \]
   \[
   \frac{dy}{dt} = -e^{-t}\sin t + e^{-t}\cos t = e^{-t}(\cos t - \sin t)
   \]

2. **Set up the arc length integral:**
   \[
   L = \int_{0}^{\frac{\pi}{2}} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt
   \]

3. **Substitute the derivatives:**
   \[
   L = \int_{0}^{\frac{\pi}{2}} \sqrt{\left(-e^{-t}(\cos t + \sin t)\right)^2 + \left(e^{-t}(\cos t - \sin t)\right)^2} \, dt
   \]

4. **Simplify the integrand:**
   \[
   L = \int_{0}^{\frac{\pi}{2}} \sqrt{e^{-2t}(\cos t + \sin t)^2 + e^{-2t}(\cos t - \sin t)^2} \, dt
   \]
   \[
   = \int_{0}^{\frac{\pi}{2}} \sqrt{e^{-2t} \left[ (\cos t + \sin t)^2 + (\cos t - \sin t)^2 \right]} \, dt
   \]

5. **Expand and combine terms:**
   \[
   (\cos t + \sin t)^2 = \cos^2 t + 2\cos t \sin t + \sin^2 t = 1 + 2\cos t \sin t
   \]
   \[
   (\cos t - \sin t)^2 = \cos^2 t - 2\cos t \sin t + \sin^2 t = 1 - 2\cos t \sin t
   \]
   \[
   (\cos t + \sin t)^2 + (\cos t - \sin t)^2 = (1 + 2\cos t \sin t) + (1 - 2\cos t \sin t) = 2
   \]

6. **Simplify the integral:**
   \[
   L = \int_{0}^{\frac{\pi}{2}} \sqrt{e^{-2t} \cdot 2} \, dt = \int_{0}^{\frac{\pi}{2}} \sqrt{2} e^{-t} \, dt
   \]

7. **Evaluate the integral:**
   \[
   L = \sqrt{2} \int_{0}^{\frac{\pi}{2}} e^{-t} \, dt = \sqrt{2} \left[ -e^{-t} \right]_{0}^{\frac{\pi}{2}}
   \]
   \[
   = \sqrt{2} \left( -e^{-\frac{\pi}{2}} + 1 \right) = \sqrt{2} \left( 1 - e^{-\frac{\pi}{2}} \right)
   \]

### Part (2): Finding the Area \( S \) of the Region Bounded by \( C \), the \( x \)-axis, and the \( y \)-axis

1. **Set up the area integral:**
   \[
   S = \int_{0}^{\frac{\pi}{2}} y \frac{dx}{dt} \, dt
   \]

2. **Substitute \( y \) and \( \frac{dx}{dt} \):**
   \[
   y = e^{-t} \sin t, \quad \frac{dx}{dt} = -e^{-t}(\cos t + \sin t)
   \]
   \[
   S = \int_{0}^{\frac{\pi}{2}} e^{-t} \sin t \cdot \left( -e^{-t}(\cos t + \sin t) \right) \, dt
   \]
   \[
   = -\int_{0}^{\frac{\pi}{2}} e^{-2t} \sin t (\cos t + \sin t) \, dt
   \]

3. **Distribute and simplify:**
   \[
   = -\int_{0}^{\frac{\pi}{2}} e^{-2t} (\sin t \cos t + \sin^2 t) \, dt
   \]

4. **Split the integral:**
   \[
   = -\int_{0}^{\frac{\pi}{2}} e^{-2t} \sin t \cos t \, dt - \int_{0}^{\frac{\pi}{2}} e^{-2t} \sin^2 t \, dt
   \]

5. **Evaluate each integral separately:**

   For the first integral:
   \[
   \int_{0}^{\frac{\pi}{2}} e^{-2t} \sin t \cos t \, dt = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} e^{-2t} \sin 2t \, dt
   \]
   Using integration by parts or a suitable substitution, this integral evaluates to zero because the integrand is an odd function over the symmetric interval.

   For the second integral:
   \[
   \int_{0}^{\frac{\pi}{2}} e^{-2t} \sin^2 t \, dt
   \]
   Using the identity \(\sin^2 t = \frac{1 - \cos 2t}{2}\):
   \[
   = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} e^{-2t} (1 - \cos 2t) \, dt
   \]
   \[
   = \frac{1}{2} \left( \int_{0}^{\frac{\pi}{2}} e^{-2t} \, dt - \int_{0}^{\frac{\pi}{2}} e^{-2t} \cos 2t \, dt \right)
   \]

6. **Evaluate the first part:**
   \[
   \int_{0}^{\frac{\pi}{2}} e^{-2t} \, dt = \left[ -\frac{1}{2} e^{-2t} \right]_{0}^{\frac{\pi}{2}} = -\frac{1}{2} \left( e^{-\pi} - 1 \right) = \frac{1 - e^{-\pi}}{2}
   \]

7. **Evaluate the second part:**
   \[
   \int_{0}^{\frac{\pi}{2}} e^{-2t} \cos 2t \, dt
   \]
   This integral also evaluates to zero because the integrand is an odd function over the symmetric interval.

8. **Combine the results:**
   \[
   S = \frac{1}{2} \left( \frac{1 - e^{-\pi}}{2} \right) = \frac{1 - e^{-\pi}}{4}
   \]

The final answer is \( \boxed{ \sqrt{2} \left( 1 - e^{-\frac{\pi}{2}} \right) } \) for the length and \( \frac{1 - e^{-\pi}}{4} \) for the area.