1. **Problem Statement:** Find the surface area generated by revolving the curve about the y-axis for each given function and interval. 2. **Formula:** The surface area $S$ when revolving a curve $x=f(y)$ from $y=a$ to $y=b$ about the y-axis is given by: $$S = 2\pi \int_a^b x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy$$ 3. **Important:** We need to find $x$, $\frac{dx}{dy}$, then substitute into the formula and integrate over the given interval. --- **Problem 5:** $x=9y+1$, $0 \leq y \leq 2$ - $\frac{dx}{dy} = 9$ - Surface area: $$S = 2\pi \int_0^2 (9y+1) \sqrt{1 + 9^2} \, dy = 2\pi \int_0^2 (9y+1) \sqrt{82} \, dy$$ - Calculate integral: $$= 2\pi \sqrt{82} \left[ \frac{9y^2}{2} + y \right]_0^2 = 2\pi \sqrt{82} \left( \frac{9 \times 4}{2} + 2 \right) = 2\pi \sqrt{82} (18 + 2) = 2\pi \sqrt{82} \times 20$$ - Final answer: $$S = 40\pi \sqrt{82}$$ --- **Problem 6:** $x = y^3$, $0 \leq y \leq 1$ - $\frac{dx}{dy} = 3y^2$ - Surface area: $$S = 2\pi \int_0^1 y^3 \sqrt{1 + (3y^2)^2} \, dy = 2\pi \int_0^1 y^3 \sqrt{1 + 9y^4} \, dy$$ - Use substitution $u = 1 + 9y^4$, $du = 36y^3 dy$: $$S = 2\pi \int_{u=1}^{10} \frac{\sqrt{u}}{36} du = \frac{\pi}{18} \int_1^{10} u^{1/2} du = \frac{\pi}{18} \left[ \frac{2}{3} u^{3/2} \right]_1^{10} = \frac{\pi}{27} (10^{3/2} - 1)$$ - Final answer: $$S = \frac{\pi}{27} (10\sqrt{10} - 1)$$ --- **Problem 7:** $x = \sqrt{9 - y^2}$, $-2 \leq y \leq 2$ - $\frac{dx}{dy} = \frac{-y}{\sqrt{9 - y^2}}$ - Surface area: $$S = 2\pi \int_{-2}^2 \sqrt{9 - y^2} \sqrt{1 + \left(\frac{-y}{\sqrt{9 - y^2}}\right)^2} dy = 2\pi \int_{-2}^2 \sqrt{9 - y^2} \sqrt{1 + \frac{y^2}{9 - y^2}} dy$$ - Simplify inside the root: $$\sqrt{\frac{9 - y^2 + y^2}{9 - y^2}} = \sqrt{\frac{9}{9 - y^2}} = \frac{3}{\sqrt{9 - y^2}}$$ - So integrand: $$\sqrt{9 - y^2} \times \frac{3}{\sqrt{9 - y^2}} = 3$$ - Integral: $$S = 2\pi \int_{-2}^2 3 \, dy = 6\pi [y]_{-2}^2 = 6\pi (2 - (-2)) = 24\pi$$ - Final answer: $$S = 24\pi$$ --- **Problem 8:** $x = 2\sqrt{1 - y}$, $-1 \leq y \leq 0$ - $\frac{dx}{dy} = 2 \times \frac{1}{2\sqrt{1 - y}} \times (-1) = -\frac{1}{\sqrt{1 - y}}$ - Surface area: $$S = 2\pi \int_{-1}^0 2\sqrt{1 - y} \sqrt{1 + \left(-\frac{1}{\sqrt{1 - y}}\right)^2} dy = 4\pi \int_{-1}^0 \sqrt{1 - y} \sqrt{1 + \frac{1}{1 - y}} dy$$ - Simplify inside root: $$\sqrt{\frac{1 - y + 1}{1 - y}} = \sqrt{\frac{2 - y}{1 - y}}$$ - So integrand: $$\sqrt{1 - y} \times \sqrt{\frac{2 - y}{1 - y}} = \sqrt{2 - y}$$ - Integral: $$S = 4\pi \int_{-1}^0 \sqrt{2 - y} \, dy$$ - Substitute $u = 2 - y$, $du = -dy$: $$S = 4\pi \int_{u=3}^{2} \sqrt{u} (-du) = 4\pi \int_2^3 u^{1/2} du = 4\pi \left[ \frac{2}{3} u^{3/2} \right]_2^3 = \frac{8\pi}{3} (3^{3/2} - 2^{3/2})$$ - Final answer: $$S = \frac{8\pi}{3} (3\sqrt{3} - 2\sqrt{2})$$