Geometric Applications of Double Integrals
Solved Problems
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 \(z = xy.\)
Figure 9. 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 \(z = 1 - x.\)
Figure 11. 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 \(13\)).
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 \(14\)).
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,
\[H - z = \frac{{Hr}}{R}\;\; \text{or}\;\;z\left( {x,y} \right) = H - \frac{{Hr}}{R} = \frac{H}{R}\left( {R - r} \right) = \frac{H}{R}\left( {R - \sqrt {{x^2} + {y^2}} } \right).\]
Figure 15. 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
\[S = 2{S_{\frac{1}{2}}} = 4\pi {a^2}.\]
