1. **State the problem:** We want to find the volume $V$ of the solid generated by rotating the region bounded by the curve $y = \frac{x^4}{2}$ and the line $y = 2$ between $x=0$ and $x=4$ about the x-axis. 2. **Identify the curves and region:** The curve is $y = \frac{x^4}{2}$ and the line is $y=2$. We consider $x$ from 0 to 4. 3. **Find intersection points:** Solve $\frac{x^4}{2} = 2$ to find where the curves intersect. $$\frac{x^4}{2} = 2 \implies x^4 = 4 \implies x = \sqrt[4]{4} = \sqrt{2} \approx 1.414$$ 4. **Set up the volume integral using the washer method:** The volume of the solid formed by rotating around the x-axis is $$V = \pi \int_0^4 \left(R(x)^2 - r(x)^2\right) dx$$ where $R(x)$ is the outer radius and $r(x)$ is the inner radius. 5. **Determine outer and inner radii in each interval:** - For $0 \leq x \leq \sqrt{2}$, the upper curve is $y=2$ and the lower curve is $y=\frac{x^4}{2}$. - For $\sqrt{2} \leq x \leq 4$, the curve $y=\frac{x^4}{2}$ is above $y=2$. 6. **Write the volume integral as sum of two integrals:** $$V = \pi \int_0^{\sqrt{2}} \left(2^2 - \left(\frac{x^4}{2}\right)^2\right) dx + \pi \int_{\sqrt{2}}^4 \left(\left(\frac{x^4}{2}\right)^2 - 2^2\right) dx$$ 7. **Simplify the integrands:** $$2^2 = 4$$ $$\left(\frac{x^4}{2}\right)^2 = \frac{x^8}{4}$$ So, $$V = \pi \int_0^{\sqrt{2}} \left(4 - \frac{x^8}{4}\right) dx + \pi \int_{\sqrt{2}}^4 \left(\frac{x^8}{4} - 4\right) dx$$ 8. **Evaluate the integrals:** Calculate each integral separately. First integral: $$\int_0^{\sqrt{2}} \left(4 - \frac{x^8}{4}\right) dx = \int_0^{\sqrt{2}} 4 dx - \int_0^{\sqrt{2}} \frac{x^8}{4} dx = 4x \Big|_0^{\sqrt{2}} - \frac{1}{4} \cdot \frac{x^9}{9} \Big|_0^{\sqrt{2}} = 4\sqrt{2} - \frac{(\sqrt{2})^9}{36}$$ Note that $(\sqrt{2})^9 = (2^{1/2})^9 = 2^{9/2} = 2^{4 + 1/2} = 2^4 \cdot 2^{1/2} = 16 \sqrt{2}$. So, $$4\sqrt{2} - \frac{16 \sqrt{2}}{36} = 4\sqrt{2} - \frac{4 \sqrt{2}}{9} = \frac{36\sqrt{2} - 4\sqrt{2}}{9} = \frac{32\sqrt{2}}{9}$$ Second integral: $$\int_{\sqrt{2}}^4 \left(\frac{x^8}{4} - 4\right) dx = \frac{1}{4} \int_{\sqrt{2}}^4 x^8 dx - 4 \int_{\sqrt{2}}^4 dx = \frac{1}{4} \cdot \frac{x^9}{9} \Big|_{\sqrt{2}}^4 - 4(x) \Big|_{\sqrt{2}}^4 = \frac{4^9 - (\sqrt{2})^9}{36} - 4(4 - \sqrt{2})$$ Calculate powers: $4^9 = (2^2)^9 = 2^{18} = 262144$ Recall $(\sqrt{2})^9 = 16 \sqrt{2}$ from above. So, $$\frac{262144 - 16 \sqrt{2}}{36} - 16 + 4\sqrt{2} = \frac{262144}{36} - \frac{16 \sqrt{2}}{36} - 16 + 4\sqrt{2} = \frac{262144}{36} - 16 + \left(4 - \frac{16}{36}\right) \sqrt{2}$$ Simplify the coefficient of $\sqrt{2}$: $$4 - \frac{16}{36} = 4 - \frac{4}{9} = \frac{36}{9} - \frac{4}{9} = \frac{32}{9}$$ So the second integral is: $$\frac{262144}{36} - 16 + \frac{32 \sqrt{2}}{9}$$ 9. **Sum both integrals and multiply by $\pi$:** $$V = \pi \left( \frac{32 \sqrt{2}}{9} + \frac{262144}{36} - 16 + \frac{32 \sqrt{2}}{9} \right) = \pi \left( \frac{64 \sqrt{2}}{9} + \frac{262144}{36} - 16 \right)$$ Convert $16$ to fraction with denominator 36: $$16 = \frac{576}{36}$$ So, $$V = \pi \left( \frac{64 \sqrt{2}}{9} + \frac{262144 - 576}{36} \right) = \pi \left( \frac{64 \sqrt{2}}{9} + \frac{261568}{36} \right)$$ Simplify $\frac{261568}{36}$ by dividing numerator and denominator by 4: $$\frac{261568}{36} = \frac{65392}{9}$$ So, $$V = \pi \left( \frac{64 \sqrt{2}}{9} + \frac{65392}{9} \right) = \pi \cdot \frac{64 \sqrt{2} + 65392}{9}$$ 10. **Answer to part (a):** The correct integral expression matches $$V = \pi \int_0^{\sqrt{2}} \left(4 - \frac{x^8}{4}\right) dx + \pi \int_{\sqrt{2}}^4 \left(\frac{x^8}{4} - 4\right) dx$$ which corresponds to the option similar to $$V = \int_0^1 \pi (4 - \frac{x^4}{4}) dx + \int_1^4 \pi (\frac{x^4}{4} - 4) dx$$ with adjusted limits and powers. The closest correct option is: "V = \int_0^1 \pi (4 - \frac{x^4}{4}) dx + \int_1^4 \pi (\frac{x^4}{4} - 4) dx" 11. **Answer to part (b):** $$V = \pi \cdot \frac{64 \sqrt{2} + 65392}{9}$$ This is the exact volume. --- "slug": "volume rotation", "subject": "calculus", "svg": "", "desmos": {"latex": "y=\frac{x^4}{2}", "features": {"intercepts": true, "extrema": true}}, "q_count": 2