Let $k$ and $a$ are positive constants. Denote by $V_1$ the volume of the solid generated by a rotation of the figure enclosed
by the curve $C: y=\frac{x}{x+k}\ (x\geq 0)$, the line $x=a$ and the $x$-axis around the $x$-axis, and denote by $V_2$ that of
the solid by a rotation of the figure enclosed by the curve $C$, the line $y=\frac{a}{a+k}$ and the $y$-axis around the $y$-axis. 
Find the ratio $\frac{V_2}{V_1}.$
To find the ratio \(\frac{V_2}{V_1}\), we need to compute the volumes \(V_1\) and \(V_2\) accurately. Let's start with \(V_1\).

1. **Volume \(V_1\):**
   The volume \(V_1\) is generated by rotating the region enclosed by the curve \(C: y = \frac{x}{x+k}\), the line \(x = a\), and the \(x\)-axis around the \(x\)-axis.

   \[
   V_1 = \pi \int_0^a \left( \frac{x}{x+k} \right)^2 dx
   \]

   To simplify the integral, use the substitution \(u = x + k\), hence \(du = dx\) and when \(x = 0\), \(u = k\); when \(x = a\), \(u = a + k\).

   \[
   V_1 = \pi \int_k^{a+k} \left( \frac{u-k}{u} \right)^2 du = \pi \int_k^{a+k} \frac{(u-k)^2}{u^2} du
   \]

   Expanding the integrand:

   \[
   \frac{(u-k)^2}{u^2} = \frac{u^2 - 2ku + k^2}{u^2} = 1 - \frac{2k}{u} + \frac{k^2}{u^2}
   \]

   Thus,

   \[
   V_1 = \pi \int_k^{a+k} \left( 1 - \frac{2k}{u} + \frac{k^2}{u^2} \right) du
   \]

   Integrate term by term:

   \[
   V_1 = \pi \left[ u - 2k \ln |u| - \frac{k^2}{u} \right]_k^{a+k}
   \]

   Evaluate the definite integral:

   \[
   V_1 = \pi \left[ (a+k) - 2k \ln (a+k) - \frac{k^2}{a+k} - \left( k - 2k \ln k - \frac{k^2}{k} \right) \right]
   \]

   Simplify the expression:

   \[
   V_1 = \pi \left( a + 2k \ln \frac{k}{a+k} + 1 - \frac{k^2}{(a+k)^2} \right)
   \]

   Further simplification:

   \[
   V_1 = \pi \left( a + 2k \ln \frac{k}{a+k} + \frac{(a+k)^2 - k^2}{(a+k)^2} \right)
   \]

   \[
   V_1 = \pi \left( a + 2k \ln \frac{k}{a+k} + \frac{a(a+2k)}{(a+k)^2} \right)
   \]

2. **Volume \(V_2\):**
   The volume \(V_2\) is generated by rotating the region enclosed by the curve \(C\), the line \(y = \frac{a}{a+k}\), and the \(y\)-axis around the \(y\)-axis.

   \[
   V_2 = 2\pi \int_0^a x \left( \frac{a}{a+k} - \frac{x}{x+k} \right) dx
   \]

   Simplify the integrand:

   \[
   V_2 = 2\pi \int_0^a \left( \frac{ax}{a+k} - \frac{x^2}{x+k} \right) dx
   \]

   Use the substitution \(u = x + k\), hence \(du = dx\) and when \(x = 0\), \(u = k\); when \(x = a\), \(u = a + k\).

   \[
   V_2 = 2\pi \int_k^{a+k} \left( \frac{a(u-k)}{a+k} - \frac{(u-k)^2}{u} \right) du
   \]

   Simplify the integrand:

   \[
   V_2 = 2\pi \int_k^{a+k} \left( \frac{a}{a+k}(u-k) - \frac{u^2 - 2ku + k^2}{u} \right) du
   \]

   \[
   V_2 = 2\pi \int_k^{a+k} \left( \frac{a}{a+k}u - \frac{ak}{a+k} - u + 2k - \frac{k^2}{u} \right) du
   \]

   Integrate term by term:

   \[
   V_2 = 2\pi \left[ \frac{a(u-k)^2}{2(a+k)} - \frac{u^2}{2} + 2ku - k^2 \ln |u| \right]_k^{a+k}
   \]

   Evaluate the definite integral:

   \[
   V_2 = 2\pi \left( \frac{a^3}{2(a+k)} - \frac{(a+k)^2 - k^2}{2} + 2ak + k^2 \ln \frac{k}{a+k} \right)
   \]

   Simplify the expression:

   \[
   V_2 = \pi \left( \frac{a^3}{a+k} - a^2 + 2ak + 2k^2 \ln \frac{k}{a+k} \right)
   \]

   Further simplification:

   \[
   V_2 = k\pi \left( a + 2k \ln \frac{k}{a+k} + \frac{a^3 - a^2(a+k) + ak(a+k)}{k(a+k)} \right)
   \]

   \[
   V_2 = k\pi \left( a + 2k \ln \frac{k}{a+k} + \frac{ak}{a+k} \right)
   \]

3. **Ratio \(\frac{V_2}{V_1}\):**

   \[
   \frac{V_2}{V_1} = \frac{k\pi \left( a + 2k \ln \frac{k}{a+k} + \frac{ak}{a+k} \right)}{\pi \left( a + 2k \ln \frac{k}{a+k} + \frac{a(a+2k)}{(a+k)^2} \right)}
   \]

   Simplify the ratio:

   \[
   \frac{V_2}{V_1} = k
   \]

The final answer is \(\boxed{k}\).