type 1
Problem 1
Find the volume of the solid bounded above by the hemisphere
\[ z = \sqrt{16 – x^2 – y^2} \]and below by the xy-plane.
Solution:
The hemisphere has radius 4, so the projection on the xy-plane is the disk \[ x^2 + y^2 \leq 16. \] Using cylindrical coordinates: \[ x = r \cos \theta, \quad y = r \sin \theta, \quad dV = r\, dz\, dr\, d\theta, \] with \[ 0 \leq r \leq 4, \quad 0 \leq \theta \leq 2\pi, \quad 0 \leq z \leq \sqrt{16 – r^2}. \] Setting up the triple integral: \[ V = \int_0^{2\pi} \int_0^4 \int_0^{\sqrt{16 – r^2}} r \, dz \, dr \, d\theta. \] Compute the inner integral: \[ \int_0^{\sqrt{16 – r^2}} r \, dz = r \sqrt{16 – r^2}. \] Compute the integral over \(r\): \[ \int_0^4 r \sqrt{16 – r^2} \, dr. \] Let \(u = 16 – r^2\), \(du = -2r \, dr\): \[ = \frac{1}{2} \int_0^{16} u^{1/2} du = \frac{1}{3} \cdot 16^{3/2} = \frac{64}{3}. \] Finally, compute the \(\theta\) integral: \[ V = \int_0^{2\pi} \frac{64}{3} d\theta = \frac{128\pi}{3}. \] \[ \boxed{V = \tfrac{128\pi}{3}} \] —Problem 2
Find the volume of the solid bounded above by
\[ z = 9 – x^2 – y^2 \]and below by
\[ z = 0. \]Solution:
The paraboloid projects onto the disk \[ x^2 + y^2 \leq 9. \] Using cylindrical coordinates: \[ 0 \leq r \leq 3, \quad 0 \leq \theta \leq 2\pi, \quad 0 \leq z \leq 9 – r^2. \] Set up the integral: \[ V = \int_0^{2\pi} \int_0^3 \int_0^{9 – r^2} r \, dz \, dr \, d\theta. \] Inner integral: \[ \int_0^{9 – r^2} r \, dz = r(9 – r^2) = 9r – r^3. \] Integral over \(r\): \[ \int_0^3 (9r – r^3) dr = \frac{81}{4}. \] Integral over \(\theta\): \[ V = \int_0^{2\pi} \frac{81}{4} d\theta = \frac{81\pi}{2}. \] \[ \boxed{V = \tfrac{81\pi}{2}} \] —Problem 3
Find the volume of the solid bounded above by
\[ z = 9 – x^2 – y^2 \]and below by
\[ z = 5. \]Solution:
Intersection curve: \[ 9 – x^2 – y^2 = 5 \quad \Rightarrow \quad x^2 + y^2 = 4. \] In cylindrical coordinates: \[ 0 \leq r \leq 2, \quad 0 \leq \theta \leq 2\pi, \quad 5 \leq z \leq 9 – r^2. \] Set up the integral: \[ V = \int_0^{2\pi} \int_0^2 \int_5^{9 – r^2} r \, dz \, dr \, d\theta. \] Inner integral: \[ \int_5^{9 – r^2} r \, dz = r(4 – r^2) = 4r – r^3. \] Integral over \(r\): \[ \int_0^2 (4r – r^3) dr = 4. \] Integral over \(\theta\): \[ V = \int_0^{2\pi} 4 \, d\theta = 8\pi. \] \[ \boxed{V = 8\pi} \] —Problem 4
Find the volume of the solid bounded above by the plane
\[ z = 6 – 2x – 3y \]and below by the xy-plane in the first octant.
Solution:
In the first octant: \[ x \geq 0, \quad y \geq 0, \quad z \geq 0. \] From \(z \geq 0\): \[ 6 – 2x – 3y \geq 0 \quad \Rightarrow \quad 2x + 3y \leq 6. \] So: \[ 0 \leq x \leq 3, \quad 0 \leq y \leq \tfrac{6 – 2x}{3}. \] Set up the triple integral: \[ V = \int_0^3 \int_0^{\tfrac{6 – 2x}{3}} \int_0^{6 – 2x – 3y} dz \, dy \, dx. \] Inner integral: \[ \int_0^{6 – 2x – 3y} dz = 6 – 2x – 3y. \] Next integral: \[ \int_0^{\tfrac{6 – 2x}{3}} (6 – 2x – 3y) dy = \tfrac{(6 – 2x)^2}{6}. \] Final integral: \[ V = \int_0^3 \frac{(6 – 2x)^2}{6} dx = 6. \] \[ \boxed{V = 6} \]type 2
Problem 5
Find the volume of the solid bounded by the paraboloids
\[ z = x^2 + y^2, \quad z = 8 – (x^2 + y^2). \]Solution:
Intersection: \[ x^2 + y^2 = 8 – (x^2 + y^2) \;\;\Rightarrow\;\; x^2 + y^2 = 4. \] Cylindrical coordinates: \[ 0 \le r \le 2, \quad 0 \le \theta \le 2\pi, \quad r^2 \le z \le 8 – r^2. \] Integral: \[ V = \int_0^{2\pi}\int_0^2 \int_{r^2}^{8-r^2} r \, dz \, dr \, d\theta. \] Inner integral: \[ \int_{r^2}^{8-r^2} r \, dz = r(8 – 2r^2) = 8r – 2r^3. \] Over \(r\): \[ \int_0^2 (8r – 2r^3) dr = \left[4r^2 – \tfrac12 r^4\right]_0^2 = 8. \] Over \(\theta\): \[ V = \int_0^{2\pi} 8 \, d\theta = 16\pi. \] \[ \boxed{V = 16\pi} \] —Problem 6
Find the volume of the solid bounded by the cone
\[ z = \sqrt{x^2 + y^2} \]and the paraboloid
\[ z = 2 – (x^2 + y^2). \]Solution:
Intersection: \[ r = 2 – r^2 \;\;\Rightarrow\;\; r^2 + r – 2 = 0 \;\;\Rightarrow\;\; r=1. \] Cylindrical coordinates: \[ 0 \le r \le 1, \quad 0 \le \theta \le 2\pi, \quad r \le z \le 2 – r^2. \] Integral: \[ V = \int_0^{2\pi}\int_0^1 \int_r^{2-r^2} r \, dz \, dr \, d\theta. \] Inner integral: \[ \int_r^{2-r^2} r \, dz = 2r – r^2 – r^3. \] Over \(r\): \[ \int_0^1 (2r – r^2 – r^3) dr = \left[r^2 – \tfrac{r^3}{3} – \tfrac{r^4}{4}\right]_0^1 = \tfrac{5}{12}. \] Over \(\theta\): \[ V = \int_0^{2\pi} \tfrac{5}{12} d\theta = \tfrac{5\pi}{6}. \] \[ \boxed{V = \tfrac{5\pi}{6}} \] —Problem 7
Find the volume of the solid bounded by
\[ z = 1 + x^2 + y^2, \quad z = 5 – (x^2 + y^2). \]Solution:
Intersection: \[ 1 + r^2 = 5 – r^2 \;\;\Rightarrow\;\; r = \sqrt{2}. \] Cylindrical coordinates: \[ 0 \le r \le \sqrt{2}, \quad 0 \le \theta \le 2\pi, \quad 1+r^2 \le z \le 5-r^2. \] Integral: \[ V = \int_0^{2\pi}\int_0^{\sqrt{2}}\int_{1+r^2}^{5-r^2} r \, dz \, dr \, d\theta. \] Inner integral: \[ \int_{1+r^2}^{5-r^2} r \, dz = 4r – 2r^3. \] Over \(r\): \[ \int_0^{\sqrt{2}} (4r – 2r^3) dr = \left[2r^2 – \tfrac{r^4}{2}\right]_0^{\sqrt{2}} = 2. \] Over \(\theta\): \[ V = \int_0^{2\pi} 2 \, d\theta = 4\pi. \] \[ \boxed{V = 4\pi} \] —Problem 8
Find the volume of the solid bounded by the cylinder
\[ x^2 + y^2 = 4, \quad z = 0, \quad z = 3. \]Solution:
This is a cylinder of radius 2 and height 3. Cylindrical coordinates: \[ 0 \le r \le 2, \quad 0 \le \theta \le 2\pi, \quad 0 \le z \le 3. \] Volume: \[ V = \int_0^{2\pi}\int_0^2\int_0^3 r \, dz \, dr \, d\theta = \int_0^{2\pi}\int_0^2 3r \, dr \, d\theta. \] Over \(r\): \[ \int_0^2 3r \, dr = \left[\tfrac{3r^2}{2}\right]_0^2 = 6. \] Over \(\theta\): \[ V = \int_0^{2\pi} 6 \, d\theta = 12\pi. \] \[ \boxed{V = 12\pi} \] —Problem 9
Find the volume of the solid enclosed by the cylinder
\[ x^2 + y^2 = 4, \quad z = 0, \quad z = 2 + x. \]Solution:
In cylindrical coordinates \(x = r\cos\theta\): \[ 0 \le r \le 2, \quad 0 \le \theta \le 2\pi, \quad 0 \le z \le 2 + r\cos\theta. \] Integral: \[ V = \int_0^{2\pi}\int_0^2\int_0^{2+r\cos\theta} r \, dz \, dr \, d\theta. \] Inner integral: \[ \int_0^{2+r\cos\theta} r \, dz = r(2 + r\cos\theta). \] Over \(r\): \[ \int_0^2 \left(2r + r^2\cos\theta\right) dr = \left[r^2 + \tfrac{r^3}{3}\cos\theta\right]_0^2 = 4 + \tfrac{8}{3}\cos\theta. \] Over \(\theta\): \[ V = \int_0^{2\pi} \left(4 + \tfrac{8}{3}\cos\theta\right) d\theta = 8\pi + 0 = 8\pi. \] \[ \boxed{V = 8\pi} \]Problem 10
Find the volume of the portion of the solid cylinder
\[ x^2 + y^2 \le 2 \]bounded above by the surface
\[ z = x^2 + y^2 \]and bounded below by the xy-plane.
Solution:
In cylindrical coordinates: \[ r^2 \le 2,\quad 0 \le r \le \sqrt{2},\quad 0 \le \theta \le 2\pi,\quad 0 \le z \le r^2, \quad dV = r \,dz\,dr\,d\theta. \] Integral: \[ V = \int_{0}^{2\pi}\!\int_{0}^{\sqrt{2}}\!\int_{0}^{r^2} r \,dz\,dr\,d\theta. \] Compute inner integral: \[ \int_{0}^{r^2} r\,dz = r\cdot r^2 = r^3. \] Then \[ \int_{0}^{\sqrt{2}} r^3 \,dr = \left[\tfrac{r^4}{4}\right]_{0}^{\sqrt{2}} = \frac{(\sqrt{2})^4}{4} = \frac{4}{4} = 1. \] Finally, \[ V = \int_{0}^{2\pi} 1\,d\theta = 2\pi. \] \[ \boxed{V = 2\pi} \] —Problem 11
Find the volume of the solid bounded by the cylinder \[ x^2 + y^2 = 1, \] above by the plane \[ z = 2, \] and below by the cone \[ z = \sqrt{x^2 + y^2}. \] Solution: Cylinder of radius 1, use polar: \[ 0 \le r \le 1,\quad 0 \le \theta \le 2\pi,\quad \sqrt{r^2}\le z \le 2, \quad dV = r\,dz\,dr\,d\theta. \] Integral: \[ V = \int_{0}^{2\pi}\!\int_{0}^{1}\!\int_{r}^{2} r \,dz\,dr\,d\theta. \] Inner integral: \[ \int_{r}^{2} r\,dz = r(2 – r) = 2r – r^2. \] Then \[ \int_{0}^{1} (2r – r^2)\,dr = \Bigl[r^2 – \tfrac{r^3}{3}\Bigr]_{0}^{1} = 1 – \tfrac{1}{3} = \tfrac{2}{3}. \] Finally, \[ V = \int_{0}^{2\pi} \tfrac{2}{3}\,d\theta = \frac{4\pi}{3}. \] \[ \boxed{V = \frac{4\pi}{3}} \] —Problem 12
Find the volume of the solid bounded above by the cone \[ z = \sqrt{x^2 + y^2} \] and below by the plane \[ z = 0, \] inside the cylinder \[ x^2 + y^2 = 4. \] Solution: Cylinder radius 2, use polar: \[ 0 \le r \le 2,\quad 0 \le \theta \le 2\pi,\quad 0 \le z \le r, \quad dV = r\,dz\,dr\,d\theta. \] Integral: \[ V = \int_{0}^{2\pi}\!\int_{0}^{2}\!\int_{0}^{r} r \,dz\,dr\,d\theta. \] Inner integral: \[ \int_{0}^{r} r\,dz = r^2. \] Then \[ \int_{0}^{2} r^2\,dr = \Bigl[\tfrac{r^3}{3}\Bigr]_{0}^{2} = \frac{8}{3}. \] Finally, \[ V = \int_{0}^{2\pi} \frac{8}{3}\,d\theta = \frac{16\pi}{3}. \] \[ \boxed{V = \frac{16\pi}{3}} \] —Problem 13
Find the volume inside the cylinder \[ x^2 + y^2 = 1 \] and between the paraboloids \[ z = x^2 + y^2 \quad\text{and}\quad z = 2 – (x^2 + y^2). \] Solution: Cylinder radius 1, polar: \[ 0 \le r \le 1,\quad 0 \le \theta \le 2\pi,\quad r^2 \le z \le 2 – r^2, \quad dV = r\,dz\,dr\,d\theta. \] Integral: \[ V = \int_{0}^{2\pi}\!\int_{0}^{1}\!\int_{r^2}^{2-r^2} r \,dz\,dr\,d\theta. \] Inner integral: \[ \int_{r^2}^{2-r^2} r\,dz = r\bigl[(2 – r^2) – r^2\bigr] = r(2 – 2r^2) = 2r – 2r^3. \] Then \[ \int_{0}^{1} (2r – 2r^3)\,dr = \Bigl[r^2 – \tfrac{r^4}{2}\Bigr]_{0}^{1} = 1 – \tfrac12 = \tfrac12. \] Finally, \[ V = \int_{0}^{2\pi} \frac12\,d\theta = \pi. \] \[ \boxed{V = \pi} \] —Problem 14
Find the volume of the solid region common to the cylinders \[ x^2 + y^2 \le 1 \quad\text{and}\quad x^2 + z^2 \le 1. \] Solution: This is a Steinmetz solid (intersection of two orthogonal cylinders). By symmetry, the full volume is \[ V = \frac{16}{3}. \] \[ \boxed{V = \frac{16}{3}} \] —Problem 15
Find the volume of the wedge bounded by the cylinder \[ x^2 + y^2 = 1, \] the plane \[ z = 0, \] and the plane \[ z = y. \] Solution: Cylindrical coordinates: \[ 0 \le r \le 1,\quad 0 \le \theta \le \pi,\quad 0 \le z \le r\sin\theta,\quad dV = r\,dz\,dr\,d\theta. \] Integral: \[ V = \int_{0}^{\pi}\!\int_{0}^{1}\!\int_{0}^{r\sin\theta} r \,dz\,dr\,d\theta = \int_{0}^{\pi}\!\int_{0}^{1} r^2\sin\theta \,dr\,d\theta. \] Integrate in \(r\): \[ \int_{0}^{1} r^2 \,dr = \tfrac{1}{3}. \] So \[ V = \tfrac{1}{3} \int_{0}^{\pi} \sin\theta \,d\theta = \tfrac{1}{3}\cdot 2 = \tfrac{2}{3}. \] \[ \boxed{V = \tfrac{2}{3}} \] # Problem 3 **Find the volume inside the cone** \[ z = 2 \sqrt{x^2 + y^2} \] **and below the plane** \[ z = 6. \] **Solution:** Using cylindrical coordinates: \[ r = \sqrt{x^2 + y^2}, \quad 0 \leq \theta \leq 2\pi, \quad 0 \leq r \leq \frac{6}{2} = 3. \] Bounds for \(z\): \[ 2r \leq z \leq 6. \] The volume integral is: \[ V = \int_0^{2\pi} \int_0^3 \int_{2r}^6 r \, dz \, dr \, d\theta. \] Step 1: Integrate with respect to \(z\): \[ \int_{2r}^6 r \, dz = r (6 – 2r) = 6r – 2 r^2. \] Step 2: Integrate with respect to \(r\): \[ \int_0^3 (6r – 2 r^2) \, dr = \left[ 3 r^2 – \frac{2 r^3}{3} \right]_0^3 = 27 – 18 = 9. \] Step 3: Integrate with respect to \(\theta\): \[ \int_0^{2\pi} 9 \, d\theta = 18 \pi. \] \[ \boxed{V = 18\pi} \] — # Problem 4 **Find the volume above the cone** \[ z = \sqrt{x^2 + y^2} \] **and inside the sphere** \[ x^2 + y^2 + z^2 = 25. \] **Solution:** Spherical coordinates: \[ 0 \leq \theta \leq 2\pi, \quad 0 \leq \phi \leq \frac{\pi}{4}, \quad 0 \leq \rho \leq 5, \] cone → \(\phi = \pi/4\). \[ V = \int_0^{2\pi} \int_0^{\pi/4} \int_0^{5} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \] Step 1: \[ \int_0^5 \rho^2 \, d\rho = \frac{125}{3} \] Step 2: \[ \int_0^{\pi/4} \sin \phi \, d\phi = 1 – \tfrac{\sqrt{2}}{2} \] Step 3: \[ \int_0^{2\pi} d\theta = 2\pi \] \[ V = \tfrac{250 \pi}{3} \Big(1 – \tfrac{\sqrt{2}}{2}\Big) \] \[ \boxed{V = \tfrac{250 \pi}{3}\Big(1 – \tfrac{\sqrt{2}}{2}\Big)} \] — # Problem 5 **Find the volume inside the double cone** \[ z^2 = x^2 + y^2, \] between \[ z=0, \quad z=4. \] **Solution:** Cylindrical coordinates: \[ 0 \leq \theta \leq 2\pi, \quad 0 \leq r \leq 4, \quad r \leq z \leq 4 \] \[ V = \int_0^{2\pi} \int_0^4 \int_r^4 r \, dz \, dr \, d\theta \] Step 1: \[ \int_r^4 r \, dz = 4r – r^2 \] Step 2: \[ \int_0^4 (4r – r^2) dr = \frac{32}{3} \] Step 3: \[ V = \frac{32}{3} \times 2\pi = \tfrac{64\pi}{3} \] \[ \boxed{V = \tfrac{64\pi}{3}} \] — # Problem 6 **Find the volume between the cone** \[ z = \sqrt{x^2 + y^2} \] **and the sphere** \[ x^2 + y^2 + z^2 = 9. \] **Solution:** \[ \phi_0 = \arctan(1) = \tfrac{\pi}{4} \] Spherical coords: \[ 0 \leq \theta \leq 2\pi, \quad 0 \leq \phi \leq \tfrac{\pi}{4}, \quad 0 \leq \rho \leq 3 \] \[ V = \int_0^{2\pi} \int_0^{\pi/4} \int_0^3 \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \] Step 1: \[ \int_0^3 \rho^2 d\rho = 9 \] Step 2: \[ \int_0^{\pi/4} \sin \phi d\phi = 1 – \tfrac{\sqrt{2}}{2} \] Step 3: \[ \int_0^{2\pi} d\theta = 2\pi \] \[ V = 18 \pi \Big(1 – \tfrac{\sqrt{2}}{2}\Big) \] \[ \boxed{V = 18 \pi \Big(1 – \tfrac{\sqrt{2}}{2}\Big)} \] — # Problem 8 **Find the volume inside the cone** \[ z = 2 \sqrt{x^2 + y^2} \] **and below the plane** \[ z = 8. \] **Solution:** \[ 0 \leq \theta \leq 2\pi, \quad 0 \leq r \leq 4, \quad 0 \leq z \leq 2r \] \[ V = \int_0^{2\pi} \int_0^4 \int_0^{2r} r \, dz \, dr \, d\theta \] Step 1: \[ \int_0^{2r} r dz = 2r^2 \] Step 2: \[ \int_0^4 2r^2 dr = \tfrac{128}{3} \] Step 3: \[ \int_0^{2\pi} d\theta = 2\pi \] \[ V = \tfrac{256\pi}{3} \] \[ \boxed{V = \tfrac{256\pi}{3}} \] — # Problem 9 **Find the volume inside the cone** \[ z = \sqrt{x^2 + y^2} \] **between the planes** \[ z = 1 \quad \text{and} \quad z = 3. \] **Solution:** \[ 0 \leq \theta \leq 2\pi, \quad 1 \leq z \leq 3, \quad 0 \leq r \leq z \] \[ V = \int_0^{2\pi} \int_1^3 \int_0^z r \, dr \, dz \, d\theta \] Step 1: \[ \int_0^z r dr = \tfrac{z^2}{2} \] Step 2: \[ \int_1^3 \tfrac{z^2}{2} dz = \tfrac{13}{3} \] Step 3: \[ \int_0^{2\pi} d\theta = 2\pi \] \[ V = \tfrac{26\pi}{3} \] \[ \boxed{V = \tfrac{26\pi}{3}} \] — # Problem 10 **Find the volume inside the cone** \[ z = 3 \sqrt{x^2 + y^2} \] **and inside the cylinder** \[ x^2 + y^2 = 4. \] **Solution:** \[ 0 \leq \theta \leq 2\pi, \quad 0 \leq r \leq 2, \quad 0 \leq z \leq 3r \] \[ V = \int_0^{2\pi} \int_0^2 \int_0^{3r} r \, dz \, dr \, d\theta \] Step 1: \[ \int_0^{3r} r dz = 3r^2 \] Step 2: \[ \int_0^2 3r^2 dr = 8 \] Step 3: \[ \int_0^{2\pi} d\theta = 2\pi \] \[ V = 16\pi \] \[ \boxed{V = 16\pi} \]Problem 16
Find the volume inside the cone \( z = 2\sqrt{x^2+y^2} \) and below the plane \( z = 6 \).
\[ V = \int_0^{2\pi} \int_0^3 \int_{2r}^6 r \, dz \, dr \, d\theta = \int_0^{2\pi} \int_0^3 (6r – 2r^2) \, dr \, d\theta = \int_0^{2\pi} 9 \, d\theta = 18\pi \] —Problem 17
Find the volume above the cone \( z = \sqrt{x^2+y^2} \) and inside the sphere \( x^2 + y^2 + z^2 = 25 \).
\[ V = \int_0^{2\pi} \int_0^{\pi/4} \int_0^5 \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta = \frac{125}{3} \cdot \left(1 – \tfrac{\sqrt{2}}{2}\right) \cdot 2\pi = \frac{250\pi}{3}\left(1 – \tfrac{\sqrt{2}}{2}\right) \] —Problem 18
Find the volume inside the double cone \( z^2 = x^2+y^2 \) between \( z=0 \) and \( z=4 \).
\[ V = \int_0^{2\pi} \int_0^4 \int_r^4 r \, dz \, dr \, d\theta = \int_0^{2\pi} \int_0^4 (4r – r^2) \, dr \, d\theta = \int_0^{2\pi} \tfrac{32}{3} \, d\theta = \tfrac{64\pi}{3} \] —Problem 19
Find the volume between the cone \( z = \sqrt{x^2+y^2} \) and the sphere \( x^2+y^2+z^2 = 9 \).
\[ V = \int_0^{2\pi} \int_0^{\pi/4} \int_0^3 \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta = 9 \cdot \left(1 – \tfrac{\sqrt{2}}{2}\right) \cdot 2\pi = 18\pi\left(1 – \tfrac{\sqrt{2}}{2}\right) \] —Problem 20
Find the volume inside the cone \( z = 2\sqrt{x^2+y^2} \) and below the plane \( z = 8 \).
\[ V = \int_0^{2\pi} \int_0^4 \int_0^{2r} r \, dz \, dr \, d\theta = \int_0^{2\pi} \int_0^4 2r^2 \, dr \, d\theta = \int_0^{2\pi} \tfrac{128}{3} \, d\theta = \tfrac{256\pi}{3} \] —Problem 21
Find the volume inside the cone \( z = \sqrt{x^2+y^2} \) between the planes \( z=1 \) and \( z=3 \).
\[ V = \int_0^{2\pi} \int_1^3 \int_0^z r \, dr \, dz \, d\theta = \int_0^{2\pi} \int_1^3 \tfrac{z^2}{2} \, dz \, d\theta = \int_0^{2\pi} \tfrac{13}{3} \, d\theta = \tfrac{26\pi}{3} \] —Problem 22
Find the volume inside the cone \( z = 3\sqrt{x^2+y^2} \) and inside the cylinder \( x^2+y^2 = 4 \).
\[ V = \int_0^{2\pi} \int_0^2 \int_0^{3r} r \, dz \, dr \, d\theta = \int_0^{2\pi} \int_0^2 3r^2 \, dr \, d\theta = \int_0^{2\pi} 8 \, d\theta = 16\pi \]