Calculus

Double Integrals

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Geometric Applications of Double Integrals

Solved Problems

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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 \(z = xy.\)

A solid bounded by the surface z=xy
Figure 9.
Region of integration bounded by the line y=a-x and coordinate axes.
Figure 10.

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 \(z = 1 - x.\)

A solid bounded by the surface x^2+y^2=1 and planes z=0, x+y=1, z=1-x.
Figure 11.
Region of integration as a circular segment
Figure 12.

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 \(13\)).

A loop of the rose
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 \(14\)).

Unit sphere
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,

\[H - z = \frac{{Hr}}{R}\;\; \text{or}\;\;z\left( {x,y} \right) = H - \frac{{Hr}}{R} = \frac{H}{R}\left( {R - r} \right) = \frac{H}{R}\left( {R - \sqrt {{x^2} + {y^2}} } \right).\]
Right circular cone
Figure 15.
Similar triangles
Figure 16.

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

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