1. **State the problem:** Find the volume of the solid formed by rotating the region in the first quadrant bounded by $y=\sqrt{9-x^2}$ and $y=x$ about the $y$-axis. 2. **Identify the curves and region:** - The curve $y=\sqrt{9-x^2}$ is the upper half of the circle $x^2 + y^2 = 9$. - The line $y=x$ intersects the circle in the first quadrant. 3. **Find the points of intersection:** Set $y=x$ equal to $y=\sqrt{9-x^2}$: $$x = \sqrt{9 - x^2}$$ Square both sides: $$x^2 = 9 - x^2$$ $$2x^2 = 9$$ $$x^2 = \frac{9}{2}$$ $$x = \frac{3}{\sqrt{2}}$$ (positive root since first quadrant) 4. **Set up the volume integral using the shell method:** When rotating about the $y$-axis, the shell radius is $x$, the height is the difference between the upper and lower curves in terms of $y$. Height = $\sqrt{9 - x^2} - x$ Volume formula by shells: $$V = 2\pi \int_0^{\frac{3}{\sqrt{2}}} x \left(\sqrt{9 - x^2} - x\right) dx$$ 5. **Split the integral:** $$V = 2\pi \left( \int_0^{\frac{3}{\sqrt{2}}} x \sqrt{9 - x^2} \, dx - \int_0^{\frac{3}{\sqrt{2}}} x^2 \, dx \right)$$ 6. **Evaluate the first integral:** Use substitution $u = 9 - x^2$, so $du = -2x dx$, or $-\frac{1}{2} du = x dx$. When $x=0$, $u=9$; when $x=\frac{3}{\sqrt{2}}$, $$u = 9 - \left(\frac{3}{\sqrt{2}}\right)^2 = 9 - \frac{9}{2} = \frac{9}{2}$$ Integral becomes: $$\int_0^{\frac{3}{\sqrt{2}}} x \sqrt{9 - x^2} dx = -\frac{1}{2} \int_9^{\frac{9}{2}} \sqrt{u} du = \frac{1}{2} \int_{\frac{9}{2}}^9 u^{1/2} du$$ Calculate: $$\frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_{\frac{9}{2}}^9 = \frac{1}{3} \left(9^{3/2} - \left(\frac{9}{2}\right)^{3/2} \right)$$ Calculate powers: $$9^{3/2} = (9^{1/2})^3 = 3^3 = 27$$ $$\left(\frac{9}{2}\right)^{3/2} = \left(\frac{9}{2}\right)^{1} \times \left(\frac{9}{2}\right)^{1/2} = \frac{9}{2} \times \frac{3}{\sqrt{2}} = \frac{27}{2\sqrt{2}}$$ So the integral is: $$\frac{1}{3} \left(27 - \frac{27}{2\sqrt{2}} \right) = 9 - \frac{9}{2\sqrt{2}}$$ 7. **Evaluate the second integral:** $$\int_0^{\frac{3}{\sqrt{2}}} x^2 dx = \left[ \frac{x^3}{3} \right]_0^{\frac{3}{\sqrt{2}}} = \frac{1}{3} \left( \frac{3}{\sqrt{2}} \right)^3 = \frac{1}{3} \times \frac{27}{2\sqrt{2}} = \frac{9}{2\sqrt{2}}$$ 8. **Combine results:** $$V = 2\pi \left( 9 - \frac{9}{2\sqrt{2}} - \frac{9}{2\sqrt{2}} \right) = 2\pi \left( 9 - \frac{9}{\sqrt{2}} \right) = 18\pi \left(1 - \frac{1}{\sqrt{2}} \right)$$ **Final answer:** $$\boxed{V = 18\pi \left(1 - \frac{1}{\sqrt{2}} \right)}$$