1. **Problem statement:** Compute the volume under the surface defined by the function $$f(s,t) = e^s \sqrt{t^3} = e^s t^{3/2}$$ over the region in the first quadrant bounded by the curve $$s = \ln t$$ for $$t=1$$ and $$t=3$$. 2. **Region description:** The region lies above the curve $$s = \ln t$$, so for each $$t$$ in $$[1,3]$$, $$s$$ ranges from $$\ln t$$ to some upper limit. Since the problem states "above the curve" and no upper bound for $$s$$ is given, we assume the volume is bounded by $$s = \ln t$$ below and extends upward to some fixed upper limit or infinity. Usually, for volume, we consider the region between $$s=\ln t$$ and $$s=0$$ or another boundary. Since no upper bound is given, we assume the volume is between $$s=\ln t$$ and $$s=0$$ (since $$\ln t \leq 0$$ for $$t \in [1,3]$$ is false, so we must clarify). Actually, $$\ln 1=0$$ and $$\ln 3 > 0$$, so $$s=\ln t$$ is increasing from 0 to about 1.0986. So the region is above the curve $$s=\ln t$$, meaning $$s \geq \ln t$$, and since the problem states "in the first quadrant", $$s,t \geq 0$$. We can take $$s$$ from $$\ln t$$ to some upper limit. Since no upper limit is given, we assume the volume is bounded by $$s=\ln t$$ and $$s=3$$ (or some fixed number). But since the problem does not specify, we assume the volume is the integral over $$s$$ from $$\ln t$$ to some fixed upper limit, say $$s=3$$. 3. **Set up the double integral:** $$\text{Volume} = \int_{t=1}^3 \int_{s=\ln t}^3 e^s t^{3/2} \, ds \, dt$$ 4. **Integrate with respect to $$s$$:** $$\int_{s=\ln t}^3 e^s t^{3/2} \, ds = t^{3/2} \int_{s=\ln t}^3 e^s \, ds = t^{3/2} \left[ e^s \right]_{s=\ln t}^3 = t^{3/2} (e^3 - e^{\ln t}) = t^{3/2} (e^3 - t)$$ 5. **Simplify the integrand:** $$t^{3/2} (e^3 - t) = e^3 t^{3/2} - t^{5/2}$$ 6. **Integrate with respect to $$t$$:** $$\int_1^3 (e^3 t^{3/2} - t^{5/2}) \, dt = e^3 \int_1^3 t^{3/2} \, dt - \int_1^3 t^{5/2} \, dt$$ 7. **Compute each integral:** $$\int t^{3/2} dt = \frac{t^{5/2}}{\frac{5}{2}} = \frac{2}{5} t^{5/2}$$ $$\int t^{5/2} dt = \frac{t^{7/2}}{\frac{7}{2}} = \frac{2}{7} t^{7/2}$$ 8. **Evaluate definite integrals:** $$e^3 \cdot \frac{2}{5} [t^{5/2}]_1^3 - \frac{2}{7} [t^{7/2}]_1^3 = e^3 \cdot \frac{2}{5} (3^{5/2} - 1) - \frac{2}{7} (3^{7/2} - 1)$$ 9. **Final answer:** $$\boxed{\text{Volume} = \frac{2 e^3}{5} (3^{5/2} - 1) - \frac{2}{7} (3^{7/2} - 1)}$$ This is the exact volume under the surface over the specified region.