1. **Problem 1: Find the area enclosed by the curves** $y = x^2$ and $y^2 = 8x$. 2. First, rewrite $y^2 = 8x$ as $y = \pm \sqrt{8x} = \pm 2\sqrt{2x}$. 3. Find the points of intersection by setting $y = x^2$ equal to $y = 2\sqrt{2x}$ (considering the positive branch since $x^2 \geq 0$): $$x^2 = 2\sqrt{2x}$$ 4. Square both sides: $$x^4 = 4 \cdot 2x = 8x$$ 5. Rearrange: $$x^4 - 8x = 0$$ 6. Factor: $$x(x^3 - 8) = 0$$ 7. Solutions: $$x = 0 \quad \text{or} \quad x^3 = 8 \Rightarrow x = 2$$ 8. The curves intersect at $x=0$ and $x=2$. 9. The area enclosed is between $x=0$ and $x=2$, bounded by $y=2\sqrt{2x}$ (upper curve) and $y=x^2$ (lower curve). 10. Area $A$ is: $$A = \int_0^2 \left(2\sqrt{2x} - x^2\right) dx$$ 11. Simplify the integrand: $$2\sqrt{2x} = 2 \cdot \sqrt{2} \cdot \sqrt{x} = 2\sqrt{2} x^{1/2}$$ 12. Compute the integral: $$A = \int_0^2 \left(2\sqrt{2} x^{1/2} - x^2\right) dx = 2\sqrt{2} \int_0^2 x^{1/2} dx - \int_0^2 x^2 dx$$ 13. Evaluate each integral: $$\int_0^2 x^{1/2} dx = \left[ \frac{2}{3} x^{3/2} \right]_0^2 = \frac{2}{3} (2)^{3/2} = \frac{2}{3} \cdot 2^{1.5} = \frac{2}{3} \cdot 2 \sqrt{2} = \frac{4 \sqrt{2}}{3}$$ $$\int_0^2 x^2 dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3}$$ 14. Substitute back: $$A = 2\sqrt{2} \cdot \frac{4 \sqrt{2}}{3} - \frac{8}{3} = \frac{8 \cdot 2}{3} - \frac{8}{3} = \frac{16}{3} - \frac{8}{3} = \frac{8}{3}$$ 15. **Area enclosed by the curves is** $\boxed{\frac{8}{3}}$. --- 16. **Problem 2: Find the volume of the solid formed by rotating the area between the curves about the x-axis.** 17. The volume $V$ is given by the washer method: $$V = \pi \int_0^2 \left( (2\sqrt{2x})^2 - (x^2)^2 \right) dx$$ 18. Simplify the integrand: $$(2\sqrt{2x})^2 = 4 \cdot 2x = 8x$$ $$(x^2)^2 = x^4$$ 19. So, $$V = \pi \int_0^2 (8x - x^4) dx$$ 20. Compute the integral: $$\int_0^2 8x dx = \left[4x^2\right]_0^2 = 4 \cdot 4 = 16$$ $$\int_0^2 x^4 dx = \left[ \frac{x^5}{5} \right]_0^2 = \frac{32}{5}$$ 21. Substitute back: $$V = \pi (16 - \frac{32}{5}) = \pi \left( \frac{80}{5} - \frac{32}{5} \right) = \pi \cdot \frac{48}{5} = \frac{48\pi}{5}$$ 22. **Volume of the solid of revolution is** $\boxed{\frac{48\pi}{5}}$. --- 23. **Problem 3: Find the area enclosed between the curves** $y = x^2$ and $y = 4 - x^2$. 24. Find points of intersection by setting: $$x^2 = 4 - x^2$$ 25. Rearrange: $$2x^2 = 4 \Rightarrow x^2 = 2 \Rightarrow x = \pm \sqrt{2}$$ 26. The curves intersect at $x = -\sqrt{2}$ and $x = \sqrt{2}$. 27. The upper curve is $y = 4 - x^2$ and the lower curve is $y = x^2$. 28. Area $A$ is: $$A = \int_{-\sqrt{2}}^{\sqrt{2}} \left(4 - x^2 - x^2\right) dx = \int_{-\sqrt{2}}^{\sqrt{2}} (4 - 2x^2) dx$$ 29. Compute the integral: $$\int (4 - 2x^2) dx = 4x - \frac{2x^3}{3}$$ 30. Evaluate from $-\sqrt{2}$ to $\sqrt{2}$: $$A = \left[4x - \frac{2x^3}{3}\right]_{-\sqrt{2}}^{\sqrt{2}} = \left(4\sqrt{2} - \frac{2(\sqrt{2})^3}{3}\right) - \left(-4\sqrt{2} + \frac{2(-\sqrt{2})^3}{3}\right)$$ 31. Simplify powers: $$(\sqrt{2})^3 = (2^{1/2})^3 = 2^{3/2} = 2 \sqrt{2}$$ 32. Substitute: $$= \left(4\sqrt{2} - \frac{2 \cdot 2 \sqrt{2}}{3}\right) - \left(-4\sqrt{2} + \frac{2 \cdot (-2 \sqrt{2})}{3}\right)$$ $$= \left(4\sqrt{2} - \frac{4 \sqrt{2}}{3}\right) - \left(-4\sqrt{2} - \frac{4 \sqrt{2}}{3}\right)$$ 33. Simplify inside parentheses: $$= \left(\frac{12\sqrt{2}}{3} - \frac{4\sqrt{2}}{3}\right) - \left(-\frac{12\sqrt{2}}{3} - \frac{4\sqrt{2}}{3}\right) = \frac{8\sqrt{2}}{3} - \left(-\frac{16\sqrt{2}}{3}\right) = \frac{8\sqrt{2}}{3} + \frac{16\sqrt{2}}{3} = \frac{24\sqrt{2}}{3} = 8\sqrt{2}$$ 34. **Area enclosed between the curves is** $\boxed{8\sqrt{2}}$.