# Geometric Applications of Double Integrals

## Solved Problems

Click or tap a problem to see the solution.

### Example 5

Find the volume of the solid bounded by $z = xy, x + y = a, z = 0.$

### Example 6

Find the volume of the solid bounded by the surfaces

$z = 0, x + y = 1, {x^2} + {y^2} = 1, z = 1 - x.$

### Example 7

Find the area of one loop of the rose defined by the equation $r = \cos 2\theta.$

### Example 8

Find the volume of the unit sphere.

### Example 9

Use polar coordinates to find the volume of a right circular cone with height $$H$$ and a circular base with radius $$R.$$

### Example 10

Calculate the surface area of a sphere of radius $$a.$$

### Example 5.

Find the volume of the solid bounded by $z = xy, x + y = a, z = 0.$

Solution.

The solid lies above the triangle $$R$$ in the $$xy$$-plane (see Figures $$9,10$$) and under the surface $$z = xy.$$

The volume of the solid is

$V = \iint\limits_R {xydxdy} = \int\limits_0^a {\left[ {\int\limits_0^{a - x} {xydy} } \right]dx} = \int\limits_0^a {\left[ {\left. {\left( {\frac{{x{y^2}}}{2}} \right)} \right|_{y = 0}^{a - x}} \right]dx} = \frac{1}{2}\int\limits_0^a {x{{\left( {a - x} \right)}^2}dx} = \frac{1}{2}\int\limits_0^a {x\left( {{a^2} - 2ax + {x^2}} \right)dx} = \frac{1}{2}\int\limits_0^a {\left( {{a^2}x - 2a{x^2} + {x^3}} \right)dx} = \frac{1}{2}\left. {\left( {{a^2} \cdot \frac{{{x^2}}}{2} - 2a \cdot \frac{{{x^3}}}{3} + \frac{{{x^4}}}{4}} \right)} \right|_0^a = \frac{1}{2}\left( {\frac{{{a^2}}}{2} - \frac{{2{a^4}}}{3} + \frac{{{a^4}}}{4}} \right) = \frac{{{a^4}}}{{24}}.$

### Example 6.

Find the volume of the solid bounded by $z = xy, x + y = a, z = 0.$

Solution.

As it can be seen from the Figures $$11$$ and $$12,$$ the $$y$$-values for $$0 \le x \le 1$$ range from $$1 - x$$ to $$\sqrt {1 - {x^2}}$$ in the region of integration $$R.$$ The top is the plane $$z = 1 - x.$$

Hence, the volume of the solid is

$V = \iint\limits_R {\left( {1 - x} \right)dxdy} = \int\limits_0^1 {\left[ {\int\limits_{1 - x}^{\sqrt {1 - {x^2}} } {\left( {1 - x} \right)dy} } \right]dx} = \int\limits_0^1 {\left[ {\left( {1 - x} \right)\left. y \right|_{1 - x}^{\sqrt {1 - {x^2}} }} \right]dx} = \int\limits_0^1 {\left( {1 - x} \right)\left( {\sqrt {1 - {x^2}} - 1 + x} \right)dx} = \int\limits_0^1 {\sqrt {1 - {x^2}} dx} - \int\limits_0^1 {x\sqrt {1 - {x^2}} dx} - \int\limits_0^1 {\left( {1 + 2x - {x^2}} \right)dx} .$

We calculate these three integrals separately.

${I_1} = \int\limits_0^1 {\sqrt {1 - {x^2}} dx} .$

Make the substitution: $$x = \sin t.$$ Then $$dx = \cos tdt.$$ As seen, $$t = 0$$ when $$x = 0$$, and $$t = {\frac{\pi }{2}}$$ at $$x = 1.$$ Hence,

${I_1} = \int\limits_0^1 {\sqrt {1 - {x^2}} dx} = \int\limits_0^{\frac{\pi }{2}} {\sqrt {1 - {{\sin }^2}t} \cos tdt} = \int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}tdt} = \int\limits_0^{\frac{\pi }{2}} {\frac{{1 + \cos 2t}}{2}dt} = \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\left( {1 + \cos 2t} \right)dt} = \frac{1}{2}\left. {\left( {t + \frac{{\sin 2t}}{2}} \right)} \right|_0^{\frac{\pi }{2}} = \frac{1}{2}\left( {\frac{\pi }{2} + \frac{{\sin \pi }}{2}} \right) = \frac{\pi }{4}.$

(Check this answer against the area of the unit circle in the first quadrant).

Calculate the second integral $${I_2} = \int\limits_0^1 {x\sqrt {1 - {x^2}} dx}$$ using change of variable. Let $$1 - {x^2} = w.$$ Then $$-2xdx = dw$$ or $$xdx = {\frac{{ - dw}}{2}}.$$ We see that $$w = 1$$ at $$x = 0,$$ and $$w = 0$$ at $$x = 1.$$ So the integral is given by

${I_2} = \int\limits_0^1 {x\sqrt {1 - {x^2}} dx} = \int\limits_1^0 {\sqrt w \left( { - \frac{{dw}}{2}} \right)} = - \frac{1}{2}\int\limits_1^0 {\sqrt w dw} = \frac{1}{2}\int\limits_0^1 {\sqrt w dw} = \frac{1}{2}\int\limits_0^1 {{w^{\frac{1}{2}}}dw} = \frac{1}{2}\left. {\left( {\frac{{2{w^{\frac{3}{2}}}}}{3}} \right)} \right|_0^1 = \frac{1}{3}.$

Finally, evaluate the last integral:

${I_3} = \int\limits_0^1 {\left( {1 - 2x + {x^2}} \right)dx} = \left. {\left( {x - {x^2} + \frac{{{x^3}}}{3}} \right)} \right|_0^1 = \cancel{1} - \cancel{1} + \frac{1}{3} = \frac{1}{3}.$

Thus, the volume is

$V = {I_1} - {I_2} - {I_3} = \frac{\pi }{4} - \frac{1}{3} - \frac{1}{3} = \frac{\pi }{4} - \frac{2}{3} \approx 0,12.$

### Example 7.

Find the area of one loop of the rose defined by the equation $r = \cos 2\theta.$

Solution.

We consider the loop in the sector $$- {\frac{\pi }{4}} \le \theta \le {\frac{\pi }{4}}$$ (Figure $$13$$).

The region of integration $$R$$ can be written in the form

$R = \left\{ {\left( {r,\theta } \right)|\;0 \le r \le \cos 2\theta ,\; - \frac{\pi }{4} \le \theta \le \frac{\pi }{4}} \right\}.$

Hence, the area of the region in polar coordinates is

$A = \iint\limits_R {rdrd\theta } = \int\limits_{ -\frac{\pi }{4}}^{\frac{\pi }{4}} {\left[ {\int\limits_0^{\cos 2\theta } {rdr} } \right]d\theta } = \int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\left[ {\left. {\left( {\frac{{{r^2}}}{2}} \right)} \right|_0^{\cos 2\theta }} \right]d\theta } = \frac{1}{2}\int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {{{\cos }^2}\left( {2\theta } \right)d\theta } = \frac{1}{2}\int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\frac{{1 + \cos 4\theta }}{2}d\theta } = \frac{1}{4}\int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\left( {1 + \cos 4\theta } \right)d\theta } = \frac{1}{4}\left. {\left( {\theta + \frac{{\sin 4\theta }}{4}} \right)} \right|_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} = \frac{1}{4}\left[ {\left( {\frac{\pi }{4} + \frac{{\sin \pi }}{4}} \right) - \left( { - \frac{\pi }{4} + \frac{{\sin \left( { - \pi } \right)}}{4}} \right)} \right] = \frac{\pi }{8}.$

### Example 8.

Find the volume of the unit sphere.

Solution.

The equation of the sphere with radius $$1$$ is $${x^2} + {y^2} + {z^2}$$ $$= 1$$ (Figure $$14$$).

Because of symmetry we find the volume of the upper hemisphere and then multiply the result by $$2.$$ The equation of the upper hemisphere is

$z = \sqrt {1 - \left( {{x^2} + {y^2}} \right)} .$

Transforming to polar coordinates, we have

$z\left( {r,\theta } \right) = \sqrt {1 - {r^2}} .$

In polar coordinates, the region of integration $$R$$ is given by the set

$R = \left\{ {\left( {r,\theta } \right)|\;0 \le r \le 1,\; 0 \le \theta \le 2\pi } \right\}.$

Hence, the volume of the upper hemisphere is

$V_{\frac{1}{2}} = \iint\limits_R {\sqrt {1 - {r^2}} rdrd\theta } = \int\limits_0^{2\pi } {d\theta } \int\limits_0^1 {\sqrt {1 - {r^2}} rdr} = 2\pi \int\limits_0^1 {\sqrt {1 - {r^2}} rdr} .$

We make the substitution to evaluate the latter integral. Let $$1 - {r^2} = t.$$ Then $$-2rdr = dt$$ or $$rdr = - {\frac{{dt}}{2}}.$$ Refine the limits of integration: $$t = 1$$ when $$r = 0$$, and $$t = 0$$ when $$r = 1.$$ Thus,

$V_{\frac{1}{2}} = 2\pi \int\limits_0^1 {\sqrt {1 - {r^2}} rdr} = 2\pi \int\limits_1^0 {\sqrt t \left( { - \frac{{dt}}{2}} \right)} = - \pi \int\limits_1^0 {\sqrt t dt} = \pi \int\limits_0^1 {{t^{\frac{1}{2}}}dt} = \pi \left. {\left( {\frac{{{t^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right)} \right|_0^1 = \frac{{2\pi }}{3}.$

Then, the volume of the unit sphere is

$V = 2{V_{\frac{1}{2}}} = \frac{{4\pi }}{3}.$

### Example 9.

Use polar coordinates to find the volume of a right circular cone with height $$H$$ and a circular base with radius $$R.$$

Solution.

We first find the equation of the cone. Using similar triangles (Figure $$16$$), we can write

$\frac{r}{R} = \frac{{H - z}}{H},\;\; \text{where}\;\;r = \sqrt {{x^2} + {y^2}} .$

Consequently,

$H - z = \frac{{Hr}}{R}\;\; \text{or}\;\;z\left( {x,y} \right) = H - \frac{{Hr}}{R} = \frac{H}{R}\left( {R - r} \right) = \frac{H}{R}\left( {R - \sqrt {{x^2} + {y^2}} } \right).$

Then the volume of the cone is

$V = \iint\limits_R {z\left( {x,y} \right)dxdy} = \iint\limits_R {\frac{H}{R}\left( {R - \sqrt {{x^2} + {y^2}} } \right)dxdy} = \frac{H}{R}\iint\limits_R {\left( {R - r} \right)rdrd\theta } = \frac{H}{R}\int\limits_0^{2\pi } {\left[ {\int\limits_0^R {\left( {R - r} \right)drd} } \right]d\theta } = \frac{H}{R}\int\limits_0^{2\pi } {d\theta } \int\limits_0^R {\left( {Rr - {r^2}} \right)dr} = \frac{{2\pi H}}{R}\int\limits_0^R {\left( {Rr - {r^2}} \right)dr} = \frac{{2\pi H}}{R}\left. {\left( {\frac{{R{r^2}}}{2} - \frac{{{r^3}}}{3}} \right)} \right|_{r = 0}^R = \frac{{2\pi H}}{R}\left( {\frac{{{R^3}}}{2} - \frac{{{R^3}}}{3}} \right) = \frac{{2\pi H}}{R} \cdot \frac{{{R^3}}}{6} = \frac{{\pi {R^2}H}}{3}.$

### Example 10.

Calculate the surface area of a sphere of radius $$a.$$

Solution.

Consider the upper hemisphere of the sphere. Its equation is

${x^2} + {y^2} + {z^2} = {a^2}\;\; \text{or}\;\;z = \sqrt {{a^2} - {x^2} - {y^2}} .$

Obviously, the region of integration $$R$$ is the disk of the same radius $$a$$ centered at the origin. The surface area of the hemisphere is given by formula

$S_{\frac{1}{2}} = \iint\limits_R {\sqrt {1 + {{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} dxdy}$

Calculate the partial derivatives:

$\frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\sqrt {{a^2} - {x^2} - {y^2}} = \frac{{ - \cancel{2}x}}{{\cancel{2}\sqrt {{a^2} - {x^2} - {y^2}} }} = - \frac{x}{z},$
$\frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\sqrt {{a^2} - {x^2} - {y^2}} = \frac{{ - \cancel{2}y}}{{\cancel{2}\sqrt {{a^2} - {x^2} - {y^2}} }} = - \frac{y}{z}.$

Substituting these derivatives, we have

$S_{\frac{1}{2}} = \iint\limits_R {\sqrt {1 + {{\left( {\frac{{\partial z}}{{\partial x}}} \right)}^2} + {{\left( {\frac{{\partial z}}{{\partial y}}} \right)}^2}} dxdy} = \iint\limits_R {\sqrt {1 + \frac{{{x^2}}}{{{z^2}}} + \frac{{{y^2}}}{{{z^2}}}} dxdy} = \iint\limits_R {\sqrt {\frac{{{z^2} + {x^2} + {y^2}}}{{{z^2}}}} dxdy} = \iint\limits_R {\frac{a}{z}dxdy} .$

Convert the double integral into polar coordinates.

$S_{\frac{1}{2}} = \iint\limits_R {\frac{a}{z}dxdy} = \int\limits_0^{2\pi } {\int\limits_0^a {\frac{a}{{\sqrt {{a^2} - {r^2}} }}rdrd\theta } } = a\int\limits_0^{2\pi } {d\theta } \int\limits_0^a {\frac{{rdr}}{{\sqrt {{a^2} - {r^2}} }}} = - 2\pi a\int\limits_0^a {\frac{{d\left( {{a^2} - {r^2}} \right)}}{{2\sqrt {{a^2} - {r^2}} }}} = - 2\pi a\left. {\left( {\sqrt {{a^2} - {r^2}} } \right)} \right|_{r = 0}^a = - 2\pi a\left( {0 - a} \right) = 2\pi {a^2}.$

So the surface area of the entire sphere is

$S = 2{S_{\frac{1}{2}}} = 4\pi {a^2}.$