1. **Problem Statement:** Find the volume under the plane $$3x + 2y - z = 0$$ and above the region enclosed by the parabolas $$y = x^2$$ and $$x = y^2$$. 2. **Understanding the region:** The curves $$y = x^2$$ and $$x = y^2$$ intersect where $$y = x^2$$ and $$x = y^2$$ imply $$y = (y^2)^2 = y^4$$, so $$y = y^4$$. This gives $$y(y^3 - 1) = 0$$, so $$y=0$$ or $$y=1$$. Correspondingly, $$x = y^2$$ gives $$x=0$$ or $$x=1$$. So the region is bounded between $0$ and $1$ in both $x$ and $y$. 3. **Setting up the integral:** The plane equation can be rewritten as $$z = 3x + 2y$$. 4. **Limits of integration:** For the region between the parabolas, solve for $x$ in terms of $y$: - From $$y = x^2$$, $$x = \sqrt{y}$$ (right curve) - From $$x = y^2$$, $$x = y^2$$ (left curve) Since $$y$$ goes from 0 to 1, for each fixed $$y$$, $$x$$ goes from $$y^2$$ to $$\sqrt{y}$$. 5. **Volume integral:** $$V = \int_0^1 \int_{x=y^2}^{x=\sqrt{y}} (3x + 2y) \, dx \, dy$$ 6. **Integrate with respect to $$x$$:** $$\int_{y^2}^{\sqrt{y}} (3x + 2y) \, dx = \left[ \frac{3}{2}x^2 + 2yx \right]_{x=y^2}^{x=\sqrt{y}} = \left( \frac{3}{2} (\sqrt{y})^2 + 2y (\sqrt{y}) \right) - \left( \frac{3}{2} (y^2)^2 + 2y (y^2) \right)$$ Simplify: $$= \left( \frac{3}{2} y + 2y^{3/2} \right) - \left( \frac{3}{2} y^4 + 2 y^3 \right) = \frac{3}{2} y + 2 y^{3/2} - \frac{3}{2} y^4 - 2 y^3$$ 7. **Integrate with respect to $$y$$:** $$V = \int_0^1 \left( \frac{3}{2} y + 2 y^{3/2} - \frac{3}{2} y^4 - 2 y^3 \right) dy = \left[ \frac{3}{4} y^2 + \frac{4}{5} y^{5/2} - \frac{3}{10} y^5 - \frac{1}{2} y^4 \right]_0^1$$ Evaluate at 1: $$= \frac{3}{4} + \frac{4}{5} - \frac{3}{10} - \frac{1}{2} = \frac{15}{20} + \frac{16}{20} - \frac{6}{20} - \frac{10}{20} = \frac{15 + 16 - 6 - 10}{20} = \frac{15}{20} = \frac{3}{4}$$ 8. **Final answer:** $$\boxed{\frac{3}{4}}$$