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Triple Integrals

# Calculation of Volumes Using Triple Integrals

Page 1
Problems 1-2
Page 2
Problems 3-8

The volume of a solid $$U$$ in Cartesian coordinates $$xyz$$ is given by

$V = \iiint\limits_U {dxdydz} .$

In cylindrical coordinates, the volume of a solid is defined by the formula

$V = \iiint\limits_U {\rho d\rho d\varphi dz} .$

In spherical coordinates, the volume of a solid is expressed as

$V = \iiint\limits_U {{\rho ^2}\sin \theta d\rho d\varphi d\theta } .$

## Solved Problems

Click on problem description to see solution.

### ✓Example 1

Find the volume of the cone of height $$H$$ and base radius $$R$$ (Figure $$1$$).

### ✓Example 2

Find the volume of the ball $${x^2} + {y^2} + {z^2} \le {R^2}.$$

### ✓Example 3

Find the volume of the tetrahedron bounded by the planes passing through the points $$A\left( {1,0,0} \right),$$ $$B\left( {0,2,0} \right),$$ $$C\left( {0,0,3} \right)$$ and the coordinate planes $$Oxy,$$ $$Oxz,$$ $$Oyz$$ (Figure $$2\text{).}$$

### ✓Example 4

Find the volume of the tetrahedron bounded by the planes $$x + y + z = 5,$$ $$x = 0,$$ $$y = 0,$$ $$z = 0.$$ (Figure $$4$$).

### ✓Example 5

Find the volume of the solid formed by two paraboloids:

${{z_1} = {x^2} + {y^2}\;\;\text{and}\;\;}\kern-0.3pt{{z_2} = 1 – {x^2} – {y^2}.}$

### ✓Example 6

Calculate the volume of the ellipsoid

$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1.$

### ✓Example 7

Find the volume of the solid bounded by the sphere $${x^2} + {y^2} + {z^2} = 6$$ and the paraboloid $${x^2} + {y^2} = z.$$

### ✓Example 8

Calculate the volume of the solid bounded by the paraboloid $$z = 2 – {x^2} – {y^2}$$ and the conic surface $$z = \sqrt {{x^2} + {y^2}} .$$

### Example 1.

Find the volume of the cone of height $$H$$ and base radius $$R$$ (Figure $$1$$).

#### Solution.

The cone is bounded by the surface $$z = {\large\frac{H}{R}\normalsize} \sqrt {{x^2} + {y^2}}$$ and the plane $$z = H$$ (see Figure $$1$$). Its volume in Cartesian coordinates is expressed by the formula

${V = \iiint\limits_U {dxdydz} } = {\int\limits_{ – R}^R {dx} \int\limits_{ – \sqrt {{R^2} – {x^2}} }^{\sqrt {{R^2} – {x^2}} } {dy} \int\limits_{\frac{H}{R}\sqrt {{x^2} + {y^2}} }^H {dz} .}$

Calculate this integral in cylindrical coordinates that range within the limits:

${0 \le \varphi \le 2\pi ,\;\;\;}\kern-0.3pt {0 \le \rho \le R,\;\;\;}\kern-0.3pt {\rho \le z \le H.}$

Figure 1.

As a result, we obtain (do not forget to include the Jacobian $$\rho$$):

$V = \int\limits_0^R {\rho d\rho } \int\limits_0^{2\pi } {d\varphi } \int\limits_{\frac{H}{R}\rho }^H {dz} .$

Then the volume of the cone is

${V = \int\limits_0^R {\rho d\rho } \int\limits_0^{2\pi } {d\varphi } \int\limits_{\frac{H}{R}\rho }^H {dz} } = {2\pi \int\limits_0^R {\rho d\rho } \int\limits_{\frac{H}{R}\rho }^H {dz} } = {2\pi \int\limits_0^R {\rho d\rho } \cdot \left[ {\left. z \right|_{z = \frac{H}{R}\rho }^{z = H}} \right] } = {2\pi \int\limits_0^R {\rho \left( {H – \frac{H}{R}\rho } \right)d\rho } } = {2\pi \int\limits_0^R {\left( {H\rho – \frac{H}{R}{\rho ^2}} \right)d\rho } } = {2\pi \left[ {\left. {\left( {\frac{{{\rho ^2}H}}{2} – \frac{{{\rho ^3}H}}{{3R}}} \right)} \right|_{\rho = 0}^{\rho = R}} \right] } = {2\pi \left( {\frac{{{R^2}H}}{2} – \frac{{{R^3}H}}{{3R}}} \right) } = {\frac{{2\pi {R^2}H}}{6} } = {\frac{{\pi {R^2}H}}{3}.}$

### Example 2.

Find the volume of the ball $${x^2} + {y^2} + {z^2} \le {R^2}.$$

#### Solution.

We calculate the volume of the part of the ball lying in the first octant $$\left( {x \ge 0,y \ge 0,z \ge 0} \right),$$ and then multiply the result by $$8.$$ This yields:

${V = 8\iiint\limits_U {dxdydz} } = {8\iiint\limits_{U’} {{\rho ^2}\sin \theta d\rho d\varphi d\theta } } = {8\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } \int\limits_0^R {{\rho ^2}d\rho } \int\limits_0^{\large\frac{\pi }{2}\normalsize} {\sin \theta d\theta } } = {8\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } \int\limits_0^R {{\rho ^2}d\rho } \cdot \left[ {\left. {\left( { – \cos \theta } \right)} \right|_0^{\large\frac{\pi }{2}\normalsize}} \right] } = {8\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } \int\limits_0^R {{\rho ^2}d\rho } \cdot}\kern0pt{ \left( { – \cos \frac{\pi }{2} + \cos 0} \right) } = {8\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } \int\limits_0^R {{\rho ^2}d\rho } \cdot 1 } = {8\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } \cdot \left[ {\left. {\left( {\frac{{{\rho ^3}}}{3}} \right)} \right|_0^R} \right] } = {\frac{{8{R^3}}}{3}\int\limits_0^{\large\frac{\pi }{2}\normalsize} {d\varphi } } = {\frac{{8{R^3}}}{3} \cdot \left[ {\left. \varphi \right|_0^{\large\frac{\pi }{2}\normalsize}} \right] } = {\frac{{8{R^3}}}{3} \cdot \frac{\pi }{2} } = {\frac{{4\pi {R^3}}}{3}.}$

As a result, we get the well-known expression for the volume of the ball of radius $$R.$$

Page 1
Problems 1-2
Page 2
Problems 3-8