Volume Cylindrical Da756D

1. **State the problem:** We want to find the volume of the solid $R$ bounded by the surfaces $$z = f(x,y) = 1 - y^2$$ and $$z = g(x,y) = x^2$$ using cylindrical coordinates. 2. **Convert to cylindrical coordinates:** Recall that $$x = r\cos\theta, \quad y = r\sin\theta, \quad z = z.$$ 3. **Rewrite the surfaces in cylindrical coordinates:** - $$z = 1 - y^2 = 1 - (r\sin\theta)^2 = 1 - r^2 \sin^2\theta$$ - $$z = x^2 = (r\cos\theta)^2 = r^2 \cos^2\theta$$ 4. **Find the region in the $r$-$\theta$ plane where the surfaces intersect:** Set $$1 - r^2 \sin^2\theta = r^2 \cos^2\theta$$ Rearranged: $$1 = r^2 \cos^2\theta + r^2 \sin^2\theta = r^2 (\cos^2\theta + \sin^2\theta) = r^2$$ Since $$\cos^2\theta + \sin^2\theta = 1$$, we get $$r^2 = 1 \implies r = 1$$ (taking the positive root since $r \geq 0$). 5. **Determine the bounds for $r$, $\theta$, and $z$:** - $$0 \leq r \leq 1$$ - $$0 \leq \theta \leq 2\pi$$ (full rotation around the $z$-axis) - $$z$$ is bounded below by $$g(r,\theta) = r^2 \cos^2\theta$$ and above by $$f(r,\theta) = 1 - r^2 \sin^2\theta$$ 6. **Set up the volume integral in cylindrical coordinates:** The volume element is $$dV = r \, dz \, dr \, d\theta$$. So, $$\text{Volume} = \int_0^{2\pi} \int_0^1 \int_{r^2 \cos^2\theta}^{1 - r^2 \sin^2\theta} r \, dz \, dr \, d\theta$$ 7. **Integrate with respect to $z$ first:** $$\int_{r^2 \cos^2\theta}^{1 - r^2 \sin^2\theta} r \, dz = r \left[ z \right]_{z = r^2 \cos^2\theta}^{z = 1 - r^2 \sin^2\theta} = r \left(1 - r^2 \sin^2\theta - r^2 \cos^2\theta \right) = r \left(1 - r^2 (\sin^2\theta + \cos^2\theta) \right) = r (1 - r^2)$$ 8. **Simplify the integral:** $$\text{Volume} = \int_0^{2\pi} \int_0^1 r (1 - r^2) \, dr \, d\theta$$ 9. **Integrate with respect to $r$:** $$\int_0^1 r (1 - r^2) \, dr = \int_0^1 (r - r^3) \, dr = \left[ \frac{r^2}{2} - \frac{r^4}{4} \right]_0^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$$ 10. **Integrate with respect to $\theta$:** $$\int_0^{2\pi} \frac{1}{4} \, d\theta = \frac{1}{4} \times 2\pi = \frac{\pi}{2}$$ **Final answer:** $$\boxed{\text{Volume} = \frac{\pi}{2}}$$ **Summary:** - The solid $R$ in cylindrical coordinates is $$R = \{(r, \theta, z) \mid 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi, r^2 \cos^2\theta \leq z \leq 1 - r^2 \sin^2\theta \}$$ - The volume of $R$ is $$\frac{\pi}{2}$$.