Let $a>1.$ Find the area $S(a)$ of the part surrounded by the curve $y=\frac{a^{4}}{\sqrt{(a^{2}-x^{2})^{3}}}\ (0\leq x\leq 1),\ x$ axis , $y$ axis and the line $x=1,$ then  when $a$ varies in the range of $a>1,$ then find the extremal value of $S(a).$
1. **Substitution**: We start by substituting \( x = a \sin t \) into the integral. This substitution simplifies the integral by transforming the limits and the integrand.
   \[
   dx = a \cos t \, dt
   \]
   When \( x = 0 \), \( t = 0 \). When \( x = 1 \), \( t = \sin^{-1} \frac{1}{a} \).

2. **Transforming the integral**: The integral for the area \( S(a) \) is given by:
   \[
   S(a) = \int_{0}^{1} \frac{a^4}{\sqrt{(a^2 - x^2)^3}} \, dx
   \]
   Substituting \( x = a \sin t \) and \( dx = a \cos t \, dt \), we get:
   \[
   S(a) = \int_{0}^{\sin^{-1} \frac{1}{a}} \frac{a^4}{\sqrt{(a^2 - a^2 \sin^2 t)^3}} \cdot a \cos t \, dt
   \]
   Simplifying the integrand:
   \[
   S(a) = \int_{0}^{\sin^{-1} \frac{1}{a}} \frac{a^5 \cos t}{\sqrt{(a^2 \cos^2 t)^3}} \, dt
   \]
   \[
   S(a) = \int_{0}^{\sin^{-1} \frac{1}{a}} \frac{a^5 \cos t}{a^3 (\cos^2 t)^{3/2}} \, dt
   \]
   \[
   S(a) = \int_{0}^{\sin^{-1} \frac{1}{a}} \frac{a^5 \cos t}{a^3 \cos^3 t} \, dt
   \]
   \[
   S(a) = \int_{0}^{\sin^{-1} \frac{1}{a}} \frac{a^2}{\cos^2 t} \, dt
   \]
   \[
   S(a) = a^2 \int_{0}^{\sin^{-1} \frac{1}{a}} \sec^2 t \, dt
   \]

3. **Evaluating the integral**: The integral of \(\sec^2 t\) is \(\tan t\):
   \[
   S(a) = a^2 \left[ \tan t \right]_{0}^{\sin^{-1} \frac{1}{a}}
   \]
   \[
   S(a) = a^2 \left( \tan \left( \sin^{-1} \frac{1}{a} \right) - \tan 0 \right)
   \]
   \[
   S(a) = a^2 \tan \left( \sin^{-1} \frac{1}{a} \right)
   \]

4. **Simplifying \(\tan(\sin^{-1} \frac{1}{a})\)**: We use the identity \(\tan(\sin^{-1} x) = \frac{x}{\sqrt{1 - x^2}}\):
   \[
   \tan \left( \sin^{-1} \frac{1}{a} \right) = \frac{\frac{1}{a}}{\sqrt{1 - \left( \frac{1}{a} \right)^2}} = \frac{\frac{1}{a}}{\sqrt{\frac{a^2 - 1}{a^2}}} = \frac{\frac{1}{a}}{\frac{\sqrt{a^2 - 1}}{a}} = \frac{1}{\sqrt{a^2 - 1}}
   \]
   Therefore:
   \[
   S(a) = a^2 \cdot \frac{1}{\sqrt{a^2 - 1}} = \frac{a^2}{\sqrt{a^2 - 1}}
   \]

5. **Finding the extremal value**: We need to find the minimum value of \( S(a) = \frac{a^2}{\sqrt{a^2 - 1}} \). Let \( f(a) = \frac{a^2}{\sqrt{a^2 - 1}} \).

6. **Taking the derivative**: We use the quotient rule to find \( f'(a) \):
   \[
   f(a) = \frac{a^2}{(a^2 - 1)^{1/2}}
   \]
   Let \( u = a^2 \) and \( v = (a^2 - 1)^{1/2} \). Then:
   \[
   f(a) = \frac{u}{v}
   \]
   \[
   f'(a) = \frac{v \cdot \frac{d}{da}(a^2) - u \cdot \frac{d}{da}((a^2 - 1)^{1/2})}{v^2}
   \]
   \[
   f'(a) = \frac{(a^2 - 1)^{1/2} \cdot 2a - a^2 \cdot \frac{1}{2}(a^2 - 1)^{-1/2} \cdot 2a}{(a^2 - 1)}
   \]
   \[
   f'(a) = \frac{2a(a^2 - 1)^{1/2} - a^2 \cdot \frac{a}{(a^2 - 1)^{1/2}}}{a^2 - 1}
   \]
   \[
   f'(a) = \frac{2a(a^2 - 1) - a^3}{(a^2 - 1)^{3/2}}
   \]
   \[
   f'(a) = \frac{2a^3 - 2a - a^3}{(a^2 - 1)^{3/2}}
   \]
   \[
   f'(a) = \frac{a^3 - 2a}{(a^2 - 1)^{3/2}}
   \]
   Setting \( f'(a) = 0 \):
   \[
   a^3 - 2a = 0
   \]
   \[
   a(a^2 - 2) = 0
   \]
   Since \( a > 1 \), we have:
   \[
   a^2 = 2 \implies a = \sqrt{2}
   \]

7. **Evaluating \( S(a) \) at \( a = \sqrt{2} \)**:
   \[
   S(\sqrt{2}) = \frac{(\sqrt{2})^2}{\sqrt{(\sqrt{2})^2 - 1}} = \frac{2}{\sqrt{2 - 1}} = \frac{2}{\sqrt{1}} = 2
   \]

8. **Conclusion**: The minimum value of \( S(a) \) is 2 when \( a = \sqrt{2} \).

The final answer is \(\boxed{2}\)