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:** The Fourier sine transform $$F_s(k)$$ of a function $$f(x)$$ is given by $$ F_s(k) = \sqrt{\frac{2}{\pi}} \int_0^\infty f(x) \sin(kx) \, dx $$ Since $$f(x)$$ is defined only on $$0 < x < a$$ and zero elsewhere, the integral limits reduce to $$0$$ to $$a$$. 3. **Apply the formula:** $$ F_s(k) = \sqrt{\frac{2}{\pi}} \int_0^a x \sin(kx) \, dx $$ 4. **Evaluate the integral:** Use integration by parts. Let $$u = x$$, so $$du = dx$$. Let $$dv = \sin(kx) dx$$, so $$v = -\frac{\cos(kx)}{k}$$. Then, $$ \int_0^a x \sin(kx) dx = \left. -\frac{x \cos(kx)}{k} \right|_0^a + \frac{1}{k} \int_0^a \cos(kx) dx $$ Evaluate the remaining integral: $$ \int_0^a \cos(kx) dx = \left. \frac{\sin(kx)}{k} \right|_0^a = \frac{\sin(ka)}{k} $$ So, $$ \int_0^a x \sin(kx) dx = -\frac{a \cos(ka)}{k} + \frac{1}{k} \cdot \frac{\sin(ka)}{k} = -\frac{a \cos(ka)}{k} + \frac{\sin(ka)}{k^2} $$ 5. **Substitute back into the transform:** $$ F_s(k) = \sqrt{\frac{2}{\pi}} \left(-\frac{a \cos(ka)}{k} + \frac{\sin(ka)}{k^2} \right) $$ 6. **Final answer:** $$ \boxed{F_s(k) = \sqrt{\frac{2}{\pi}} \left( \frac{\sin(ka)}{k^2} - \frac{a \cos(ka)}{k} \right)} $$ This is the Fourier sine transform of $$f(x) = x$$ on $$0 < x < a$$.