Let's solve the problem step by step: --- ### **Given:** - Sphere equation: \(x^2 + y^2 + z^2 = 9\) (radius \(R = 3\)) - Cylindrical hole: along the \(z\)-axis, radius \(a = 2\) - Find **remaining volume** using **double integral in polar coordinates** --- ## **Step 1: Visualize the Region** - The sphere is centered at the origin. - The cylinder is drilled along the \(z\)-axis, removing all points where \(x^2 + y^2 < 4\). --- ## **Step 2: Limits for \(z\)** At any point \((x, y)\), the bounds for \(z\) within the sphere are: \[ z = \pm \sqrt{9 - x^2 - y^2} \] --- ## **Step 3: Set Up in Polar Coordinates** Let: - \(x = r\cos\theta\) - \(y = r\sin\theta\) - \(r^2 = x^2 + y^2\) The cylinder removes \(r < 2\), so: - **Region of integration:** \(2 \leq r \leq 3\), \(0 \leq \theta \leq 2\pi\) --- ## **Step 4: Write the Volume Integral** The volume element is \(dz\, r\, dr\, d\theta\): \[ V = \iint_{\text{region}} \int_{-\sqrt{9 - r^2}}^{\sqrt{9 - r^2}} dz\, r\, dr\, d\theta \] Integrate \(z\) first: \[ \int_{-\sqrt{9 - r^2}}^{\sqrt{9 - r^2}} dz = 2\sqrt{9 - r^2} \] So, \[ V = \int_{0}^{2\pi} \int_{2}^{3} 2\sqrt{9 - r^2}\, r\, dr\, d\theta \] --- ## **Step 5: Integrate with respect to \(r\)** \[ V = 2 \int_{0}^{2\pi} d\theta \int_{2}^{3} r \sqrt{9 - r^2} dr \] Let’s solve the inner integral: Let \(u = 9 - r^2\) Then \(du = -2r dr\) \(\Rightarrow -\frac{1}{2} du = r dr\) When \(r = 2\), \(u = 9 - 4 = 5\) When \(r = 3\), \(u = 0\) So, \[ \int_{2}^{3} r \sqrt{9 - r^2} dr = -\frac{1}{2} \int_{5}^{0} u^{1/2} du = \frac{1}{2} \int_{0}^{5} u^{1/2} du \] \[ = \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_{0}^{5} = \frac{1}{2} \times \frac{2}{3} \left(5^{3/2} - 0\right) = \frac{1}{3} \times 5^{3/2} \] \[ 5^{3/2} = 5 \times \sqrt{5} \] So, \[ \int_{2}^{3} r \sqrt{9 - r^2} dr = \frac{1}{3} \cdot 5\sqrt{5} = \frac{5\sqrt{5}}{3} \] --- ## **Step 6: Integrate with respect to \(\theta\)** \[ V = 2 \int_{0}^{2\pi} \frac{5\sqrt{5}}{3} d\theta = 2 \cdot \frac{5\sqrt{5}}{3} \cdot 2\pi = \frac{20\pi\sqrt{5}}{3} \] --- ## **Final Answer** \[ \boxed{V = \frac{20\pi\sqrt{5}}{3}} \] This is the **volume of the remaining part of the sphere** after drilling the cylindrical hole, using double integration in polar coordinates. --- **Summary of Steps:** 1. Set up bounds in polar coordinates, removing the cylindrical hole. 2. Express limits for \(z\), integrate with respect to \(z\). 3. Integrate with respect to \(r\) using substitution. 4. Complete the integration with respect to \(\theta\). 5. Arrive at the final answer.