1. **State the problem:** Find the Fourier sine transform of the function $$f(x) = x$$ defined on the interval $$0 < x < a$$. 2. **Recall the formula for the Fourier sine transform:** $$F_s(k) = \sqrt{\frac{2}{\pi}} \int_0^a f(x) \sin(kx) \, dx$$ 3. **Substitute the given function:** $$F_s(k) = \sqrt{\frac{2}{\pi}} \int_0^a x \sin(kx) \, dx$$ 4. **Use integration by parts:** Let $$u = x$$ and $$dv = \sin(kx) dx$$. Then, $$du = dx$$ and $$v = -\frac{\cos(kx)}{k}$$. 5. **Apply integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ 6. **Calculate:** $$\int_0^a x \sin(kx) \, dx = \left[-\frac{x \cos(kx)}{k}\right]_0^a + \frac{1}{k} \int_0^a \cos(kx) \, dx$$ 7. **Evaluate the remaining integral:** $$\int_0^a \cos(kx) \, dx = \left[\frac{\sin(kx)}{k}\right]_0^a = \frac{\sin(ka)}{k}$$ 8. **Substitute back:** $$\int_0^a x \sin(kx) \, dx = -\frac{a \cos(ka)}{k} + \frac{1}{k} \cdot \frac{\sin(ka)}{k} = -\frac{a \cos(ka)}{k} + \frac{\sin(ka)}{k^2}$$ 9. **Write the final Fourier sine transform:** $$F_s(k) = \sqrt{\frac{2}{\pi}} \left(-\frac{a \cos(ka)}{k} + \frac{\sin(ka)}{k^2}\right)$$ This is the Fourier sine transform of $$f(x) = x$$ on $$0 < x < a$$.