1. **Problem statement:** Find the volume of the solid generated by revolving the region bounded by the parabola $x = y^2 + 1$ (left boundary) and the line $x = 5$ (right boundary) about three different axes: the x-axis, the y-axis, and the line $x = 5$. 2. **Identify the region:** The region is between $x = y^2 + 1$ and $x = 5$. Since $x = y^2 + 1$, the parabola opens rightward with vertex at $(1,0)$. The line $x=5$ is vertical. 3. **Find the y-limits:** Solve $y^2 + 1 = 5$ to find intersection points: $$y^2 = 4 \implies y = \pm 2$$ So the region extends from $y = -2$ to $y = 2$. --- ### Volume about the x-axis 4. **Method:** Use the shell method since the axis of rotation is horizontal (x-axis) and the region is described in terms of $y$. 5. **Shell radius:** Distance from $y$ to x-axis is $|y|$. 6. **Shell height:** Horizontal distance between $x=5$ and $x = y^2 + 1$ is $$5 - (y^2 + 1) = 4 - y^2$$ 7. **Volume integral:** $$V = 2\pi \int_{-2}^2 |y| (4 - y^2) dy$$ Since the function is even in $y$, simplify: $$V = 2\pi \times 2 \int_0^2 y (4 - y^2) dy = 4\pi \int_0^2 (4y - y^3) dy$$ 8. **Calculate integral:** $$\int_0^2 (4y - y^3) dy = \left[2y^2 - \frac{y^4}{4}\right]_0^2 = \left(2 \times 4 - \frac{16}{4}\right) - 0 = 8 - 4 = 4$$ 9. **Volume:** $$V = 4\pi \times 4 = 16\pi$$ --- ### Volume about the y-axis 10. **Method:** Use the washer method since the axis is vertical (y-axis). 11. **Bounds for $x$:** From $x = y^2 + 1$ to $x = 5$. 12. **Radius of washers:** Distance from y-axis to the curve, so outer radius $R = 5$, inner radius $r = y^2 + 1$. 13. **Volume integral:** $$V = \pi \int_{-2}^2 \left(R^2 - r^2\right) dy = \pi \int_{-2}^2 \left(5^2 - (y^2 + 1)^2\right) dy$$ 14. **Simplify integrand:** $$(y^2 + 1)^2 = y^4 + 2y^2 + 1$$ So integrand is: $$25 - (y^4 + 2y^2 + 1) = 24 - 2y^2 - y^4$$ 15. **Integral:** $$V = \pi \int_{-2}^2 (24 - 2y^2 - y^4) dy$$ Since integrand is even, double integral from 0 to 2: $$V = 2\pi \int_0^2 (24 - 2y^2 - y^4) dy$$ 16. **Calculate integral:** $$\int_0^2 (24 - 2y^2 - y^4) dy = \left[24y - \frac{2y^3}{3} - \frac{y^5}{5}\right]_0^2 = 24 \times 2 - \frac{2 \times 8}{3} - \frac{32}{5} = 48 - \frac{16}{3} - \frac{32}{5}$$ 17. **Find common denominator 15:** $$48 = \frac{720}{15}, \quad \frac{16}{3} = \frac{80}{15}, \quad \frac{32}{5} = \frac{96}{15}$$ 18. **Sum:** $$\frac{720}{15} - \frac{80}{15} - \frac{96}{15} = \frac{720 - 80 - 96}{15} = \frac{544}{15}$$ 19. **Volume:** $$V = 2\pi \times \frac{544}{15} = \frac{1088\pi}{15}$$ --- ### Volume about the line $x=5$ 20. **Method:** Use shell method with shells parallel to y-axis. 21. **Shell radius:** Distance from $x$ to line $x=5$ is $5 - x$. 22. **Shell height:** Vertical distance between $y = -\sqrt{x-1}$ and $y = \sqrt{x-1}$ is $$2\sqrt{x-1}$$ 23. **Bounds for $x$:** From $x=1$ (vertex of parabola) to $x=5$. 24. **Volume integral:** $$V = 2\pi \int_1^5 (5 - x) \times 2\sqrt{x-1} dx = 4\pi \int_1^5 (5 - x) \sqrt{x-1} dx$$ 25. **Substitute:** Let $u = x - 1$, so when $x=1$, $u=0$; when $x=5$, $u=4$. 26. **Rewrite integral:** $$4\pi \int_0^4 (5 - (u+1)) \sqrt{u} du = 4\pi \int_0^4 (4 - u) u^{1/2} du = 4\pi \int_0^4 (4u^{1/2} - u^{3/2}) du$$ 27. **Integrate:** $$\int_0^4 4u^{1/2} du = 4 \times \frac{2}{3} u^{3/2} \Big|_0^4 = \frac{8}{3} \times 8 = \frac{64}{3}$$ $$\int_0^4 u^{3/2} du = \frac{2}{5} u^{5/2} \Big|_0^4 = \frac{2}{5} \times 32 = \frac{64}{5}$$ 28. **Combine:** $$\int_0^4 (4u^{1/2} - u^{3/2}) du = \frac{64}{3} - \frac{64}{5} = \frac{320}{15} - \frac{192}{15} = \frac{128}{15}$$ 29. **Volume:** $$V = 4\pi \times \frac{128}{15} = \frac{512\pi}{15}$$ --- **Final answers:** - Volume about x-axis: $16\pi$ - Volume about y-axis: $\frac{1088\pi}{15}$ - Volume about line $x=5$: $\frac{512\pi}{15}$