1. Find the volume of the solid formed by revolving the region bounded by $y=\sqrt{3}-x$ and the x-axis from $x=0$ to $x=3$ about the x-axis. 2. Find the volume of the solid formed by revolving the region bounded by $y=x$ and $y=2-x^2$ between their intersection points about the x-axis. 3. Find the volume of the solid formed by revolving the triangular region under $y=3-2x$ from $x=0$ to $x=2$ about the x-axis. 4. Find the volume of the solid formed by revolving the region under $y=\frac{1}{x}$ from $x=1$ to $x=2$ about the x-axis. 5. Find the volume of the solid formed by revolving the region under $y=\sqrt{\cos x}$ from $x=\frac{\pi}{4}$ to $x=\frac{\pi}{2}$ about the x-axis. 6. Find the volume of the solid formed by revolving the region bounded between $y=x^2$ and $y=x^3$ from $x=0$ to $x=1$ about the x-axis. 7. Find the volume of the solid formed by revolving the region bounded by $x=\sqrt{1+y}$, the y-axis, and $y=0$ to $y=3$ about the y-axis. 8. Find the volume of the solid formed by revolving the region above $y=x^2-1$ and below $y=3$ from $x=1$ to $x=2$ about the x-axis. --- **General formula for volume by revolution about x-axis:** $$V=\pi \int_a^b [f(x)]^2 dx$$ **For revolution about y-axis:** $$V=\pi \int_c^d [g(y)]^2 dy$$ --- ### 1. Volume for $y=\sqrt{3}-x$ about x-axis from 0 to 3 $$V=\pi \int_0^3 (\sqrt{3}-x)^2 dx=\pi \int_0^3 (3 - 2\sqrt{3}x + x^2) dx$$ Calculate integral: $$=\pi \left[3x - \sqrt{3}x^2 + \frac{x^3}{3}\right]_0^3=\pi \left(9 - 9 + 9\right)=9\pi$$ --- ### 2. Volume between $y=x$ and $y=2-x^2$ about x-axis Find intersection points: $$x=2 - x^2 \Rightarrow x^2 + x - 2=0 \Rightarrow (x+2)(x-1)=0 \Rightarrow x=1, -2$$ Volume: $$V=\pi \int_{-2}^1 [(2 - x^2)^2 - x^2] dx$$ Expand: $$(2 - x^2)^2 = 4 - 4x^2 + x^4$$ So integrand: $$4 - 4x^2 + x^4 - x^2 = 4 - 5x^2 + x^4$$ Integral: $$\pi \int_{-2}^1 (4 - 5x^2 + x^4) dx = \pi \left[4x - \frac{5x^3}{3} + \frac{x^5}{5}\right]_{-2}^1$$ Calculate: At 1: $4(1) - \frac{5(1)}{3} + \frac{1}{5} = 4 - \frac{5}{3} + \frac{1}{5} = \frac{60 - 25 + 3}{15} = \frac{38}{15}$ At -2: $4(-2) - \frac{5(-2)^3}{3} + \frac{(-2)^5}{5} = -8 - \frac{5(-8)}{3} - \frac{32}{5} = -8 + \frac{40}{3} - \frac{32}{5} = \frac{-120 + 200 - 96}{15} = \frac{-16}{15}$ Difference: $$\frac{38}{15} - \left(-\frac{16}{15}\right) = \frac{54}{15} = \frac{18}{5}$$ Volume: $$V=\pi \times \frac{18}{5} = \frac{18\pi}{5}$$ --- ### 3. Volume for $y=3-2x$ from 0 to 2 about x-axis $$V=\pi \int_0^2 (3 - 2x)^2 dx = \pi \int_0^2 (9 - 12x + 4x^2) dx$$ Integral: $$=\pi \left[9x - 6x^2 + \frac{4x^3}{3}\right]_0^2 = \pi (18 - 24 + \frac{32}{3}) = \pi \left(-6 + \frac{32}{3}\right) = \pi \frac{-18 + 32}{3} = \frac{14\pi}{3}$$ --- ### 4. Volume for $y=\frac{1}{x}$ from 1 to 2 about x-axis $$V=\pi \int_1^2 \left(\frac{1}{x}\right)^2 dx = \pi \int_1^2 \frac{1}{x^2} dx = \pi \left[-\frac{1}{x}\right]_1^2 = \pi \left(-\frac{1}{2} + 1\right) = \frac{\pi}{2}$$ --- ### 5. Volume for $y=\sqrt{\cos x}$ from $\frac{\pi}{4}$ to $\frac{\pi}{2}$ about x-axis $$V=\pi \int_{\pi/4}^{\pi/2} (\sqrt{\cos x})^2 dx = \pi \int_{\pi/4}^{\pi/2} \cos x dx = \pi [\sin x]_{\pi/4}^{\pi/2} = \pi \left(1 - \frac{\sqrt{2}}{2}\right)$$ --- ### 6. Volume between $y=x^2$ and $y=x^3$ from 0 to 1 about x-axis $$V=\pi \int_0^1 [(x^2)^2 - (x^3)^2] dx = \pi \int_0^1 (x^4 - x^6) dx = \pi \left[\frac{x^5}{5} - \frac{x^7}{7}\right]_0^1 = \pi \left(\frac{1}{5} - \frac{1}{7}\right) = \pi \frac{2}{35} = \frac{2\pi}{35}$$ --- ### 7. Volume for $x=\sqrt{1+y}$ from $y=0$ to $y=3$ about y-axis Rewrite: $$x = (1+y)^{1/2}$$ Volume about y-axis: $$V=\pi \int_0^3 (1+y) dy = \pi \left[ y + \frac{y^2}{2} \right]_0^3 = \pi \left(3 + \frac{9}{2}\right) = \pi \frac{15}{2} = \frac{15\pi}{2}$$ --- ### 8. Volume above $y=x^2-1$ and below $y=3$ from $x=1$ to $x=2$ about x-axis Height of region: $$3 - (x^2 - 1) = 4 - x^2$$ Volume: $$V=\pi \int_1^2 (4 - x^2)^2 dx = \pi \int_1^2 (16 - 8x^2 + x^4) dx$$ Integral: $$=\pi \left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]_1^2 = \pi \left(32 - \frac{64}{3} + \frac{32}{5} - 16 + \frac{8}{3} - \frac{1}{5}\right)$$ Simplify inside: $$=\pi \left(16 - \frac{56}{3} + \frac{31}{5}\right) = \pi \left(16 - 18.6667 + 6.2\right) = \pi 3.5333 = \frac{53\pi}{15}$$ --- **Final answers:** 1. $9\pi$ 2. $\frac{18\pi}{5}$ 3. $\frac{14\pi}{3}$ 4. $\frac{\pi}{2}$ 5. $\pi \left(1 - \frac{\sqrt{2}}{2}\right)$ 6. $\frac{2\pi}{35}$ 7. $\frac{15\pi}{2}$ 8. $\frac{53\pi}{15}$