Examples of applications of double integrals to calculate volumes and areas are presented along with their detailed solutions.
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Calculate Volume Using Double
The volume \( V \) of the solid that is above region \( R \) in the \( xy\)-plane and below the surface \( z = f(x,y) \), given that \( f(x,y) \ge 0 \), is given by the double integral [6]
\[ \displaystyle V = \iint_R f(x,y) \;dy \;dx \]
Example 1
Use the double integral to calculate the volume of the solid that that lies between the plane \( z = 0 \) and the paraboloid \( z = 4 - x^2 - y^2 \).
Solution to Example 1
The solid between the the plane \( z = 0 \) and the parabolid is shown below.
Fig.1 - Volume Between the Plane \( z = 0 \) and the Paraboloid \( z = 4 - x^2 - y^2 \)
The region \( R \) of integration is the intersection of the paraboloid and the plane \( z =0 \) which is obtained by setting \( z = 0 \) in the equation of the paraboloid to obtain
\( 0 = 4 - x^2 - y^2 \)
Rewrite in standard form
\( x^2 + y^2 = 2^2 \)
which is a circle on the \( x y \) plane (or \( z = 0 \) ) with center at \( (0,0) \) and a radius equal to \( 2 \)
Fig.2 - Region \( R \) of integration which a circle with center at \( (0,0) \) and a radius equal to \( 2 \)
Since the region of integration is a circle, it is more efficient to use double integrals in polar coordinates.
The volume \( V \) is calculated using the double integral
\( V = \displaystyle \iint_R f(x,y) \;dy \;dx \) , where \( R \) is the region of integration defined by
\( R: 0\le \theta \le 2\pi , 0 \le r \le 2 \)
which is the circle in Fig.2 and \( f(x,y) = z = 4 - x^2 - y^2 \)
In polar coordinate, the volume is given by
\( \displaystyle V = \int_0^{2\pi} \int_0^2 (4-r^2) r dr d\theta \)
\( \displaystyle V = \int_0^{2\pi} \int_0^2 (4-r^2) r dr d\theta \)
\( = \int_0^{2\pi} \left[2r^2-\frac{r^4}{4}\right]_0^2 d\theta \)
\( = \int_0^{2\pi} ( 4 ) d\theta \)
\( = 8 \pi \)
Example 2
Find the constant \( a \) (see diagram below) so that the volume of the solid that lies above the region \( R \) located in the \( x y \) plane shown below and the surface defined by the equation \( z = \sqrt{1-x^2-y^2} \) is equal to \( 1 \) cubic unit.
Fig.3 - Given region of integrationSolution to Example 2
We first convert the given region in Fig. 2 into polar coordinates. The outer half circle \( y = \sqrt {1-x^2} \) has radius equal to \( 1 \) and the inner \( y = \sqrt {a^2-x^2} \) half circle has radius \( a \) to be found.
Fig.4 - Region of integration in polar coordinates
The region of integration in polar coordinates may be defined as
\( R: 0\le \theta \le \pi , a \le r \le 1 \)
In poplar coordinates
\( z = \sqrt{1-x^2-y^2} = \sqrt{1-r^2} \)
In polar coordinate, the volume is given by
\( \displaystyle V = \int_0^{\pi} \int_a^1 \sqrt{1-r^2} r dr d\theta \)
\( \displaystyle = \int_0^{\pi} \dfrac{\left(-a^2+1\right)^{\frac{3}{2}}}{3} d\theta \)
\( = \dfrac{\left(-a^2+1\right)^{\frac{3}{2}}}{3} \pi \)
We want the volume \( V \) to be equal to \( 1 \), hence
\( \dfrac{\left(-a^2+1\right)^{\frac{3}{2}}}{3} \pi = 1 \)
Solve the above for \( a \)
\( a = \sqrt{1-\sqrt[3]{\dfrac{9}{\pi ^2}}} \approx 0.174 \)
Example 3
Find \( a \) so that the volume of the tetrahedron bounded by the planes \( 2x + 2 y + z = a , a \gt 0\), \( y = 2 x \) , \( y = 0\) and \( z = 0 \) is equal to \( 10 \) cubic units.
Solution to Example 3
Below are graphed the planes that define the solid whose volume is to be calculated.
Fig.5 - Planes that define the solid
First find the region of integration which is a triangle defined by three vertices: \( O , B \) and \( C \)
\( O \) is the origin of the system of axis.
\( B \) is the point of intersection and the \( x\)-axis and is found by setting \( z = 0\) and \( y = 0 \) in the equation of the given plane \( 2x + 2 y + z = a \).
\( 2x = a \)
Solve for \( x \)
\( x = \dfrac{a}{2} \)
\( C \) is the point of intersection of the planes \( 2x + 2 y + z = a \) and \( y = 2 x \) and is located on the \( x y \) plane. A point on the \( x y \) plane has \( z = 0 \).
Point \( C \) is obtained by setting \( z = 0 \) in the equation \( 2x + 2 y + z = a \) solving the equation obtained
\( 2x + 2 y = a \) and \( y = 2 x \)
Solve the above by substitution to obtain
\( y = a/3 \) and \( x = a/6 \)
Hence the region \( R \) of integration on the \( x y \) plane is a triangle with vertices
\(O(0,0)\) , \( B(a/2,0) \) and \(C(a/6,a/3)\)
Fig.6 - Region of integration
The volume \( V \) of the tetrahedron is given by
\( V = \displaystyle \iint_R f(x,y) \;dy \;dx \) , where \( R \) is the region of integration defined by
\( R: \dfrac{y}{2} \le x \le \ \dfrac{a}{2} - y , 0 \le y \le \dfrac{a}{3} \)
and
\( f(x,y) = z = a - 2x - 2 y \)
\( V = \displaystyle \int_0^{\frac{a}{3}} \int_{\frac{y}{2}}^{ \frac{a}{2} - y} (a - 2x - 2 y) \;dx \;dy \)
Calculate the inner integral
\( V = \displaystyle \int_0^{\frac{a}{3}} \left[ax-2yx-x^2\right]_{\frac{y}{2}}^{ \frac{a}{2} - y} \;dy \)
\( V = \displaystyle \int_0^{\frac{a}{3}} \left(\frac{9y^2-6ay+a^2}{4}\right) \;dy \)
Evaluate the outer integral
\( V = \left[ \frac{1}{4} \left(3y^3-3a y^2+a^2 y \right) \right]_0^{\frac{a}{3}} \)
\( V = \frac{a^3}{36} \)
Note: You may check that the result above corresponds to the one that what may be obtained using the very well known formula for the volume of the tetrahedron
\( V = \dfrac{1}{3} A \times H \)
where \( A \) is the area of the base and \( H \) is the height from the base to the apex of the tetrahedron.
You may check this if you wish.
For the volume to be equal to \( 10 \)
\( \frac{a^3}{36} = 10 \)
Solve for \( a \)
\( a = \sqrt[3] {360} \approx \:7.11378 \)
Calculate Area of Surface Using Double
The area \( A \) of a region \( R \) in the plane \( x y\) is given by the double integral [6]
\[ A = \displaystyle \iint_R 1 \;dy \;dx \]
Example 4
Use double integral to calculate the area of the region bounded by the curves \( y = x^2 \) and \( y = - (x-2)^2 +4 \)
Solution to Example 4
We first plot the two curves and define the region bounded by the two curves.
Fig.7 - Region of integration
The two curves intersect at the points that are solutions to the system of equations of the two curves
\( y = x^2 \) and \( y = - (x-2)^2 +4 \)
which can be solved by substitution
\( x^2 = - (x-2)^2 +4 \)
Expand and simplify
\( 2 x^2 - 4x = 0 \)
\( 2x(x-4) = 0 \)
The above equations has two solutions
\( x = 0 \) and \( x = 4 \)
Use \( y = x^2 \) to calculate the \( y \)coordinate and hence the points
\((0,0) \) and \((2,4) \)
The region \( R \) enclosed by the two curves may be defined as follows
\( R: 0 \le x \le 2 , x^2 \le y \le - (x-2)^2 +4 \)
The area \( A \) is given by
\( A = \displaystyle \int_0^2 \int_{x^2}^{- (x-2)^2 +4} 1 \;dy \;dx \)
Calculate the inner integral
\( A = \displaystyle \int_0^2 \left[y \right]_{x^2}^{- (x-2)^2 +4} \;dx \)
Simplify
\( A = \displaystyle \int_0^2 (-2x^2+4x) \;dx \)
\( A = \left[-\frac{2x^3}{3}+2x^2\right]_0^2 = 8/3 \)
Example 5
Use double integral to calculate the area of the region common to the two circles having equations \( x^2 + (y-2)^2 = 9 \) and \( x^2 + y^2 = 9 \)
Solution to Example 5
We first plot the two curves and define the region bounded by the two curves. The region common to both circles is in light blue.
Fig.7 - Common region to the two circles
The points of intersection are found by solving the system of equations
\( x^2 + (y-2)^2 = 9 \) and \( x^2 + y^2 = 9 \)
Expand the equation on the left
\( x^2 + y^2 - 4 y + 4 = 9 \)
\( x^2 + y^2 = 9 \)
Subtract the equations
\( - 4 y + 4 = 0 \)
Solve for y
\( y = 1 \)
Substitute \( y \) by \( 1 \) in any of the equations and solve for \( x \) to obtain
\( x = \pm 2 \sqrt 2 \)
We may now define region \( R \) as
\( R: -2\sqrt 2 \le x \le 2\sqrt 2 , 2 - \sqrt{9-x^2} \le y \le \sqrt{9-x^2} \)
The area of the region \( R \) is given by
\( A = \displaystyle \int_{-2\sqrt 2}^{2\sqrt 2} \int_{2 - \sqrt{9-x^2}}^{\sqrt{9-x^2}} 1 \;dy \;dx \)
Calculate the inner integral
\( A = \displaystyle \int_{-2\sqrt 2}^{2\sqrt 2} \; \left[ y \right]_{2 - \sqrt{9-x^2}}^{\sqrt{9-x^2}} \; \;dx \)
\( A = \displaystyle \int_{-2\sqrt 2}^{2\sqrt 2} \left( 2 \sqrt{9-x^2} - 2 \right) \;dx \)
\( A = \left[ x\sqrt{9-x^2} + 9\arcsin(x/3) - 2x \right]_{-2\sqrt 2}^{2\sqrt 2} \)
Note: that details of the integral \( \int \sqrt{9-x^2}dx \) is included in appendix A.
Evaluate \( A \)
\( A = 18\arcsin \left(\frac{2\sqrt{2}}{3}\right)-4\sqrt{2} \approx 16.5 \)
Appendices
Appendix A
Calculate the integral
\( \displaystyle I = \int \sqrt{9-x^2}dx \)
Substitute
\( x = 3 \sin u \) which gives \( u = \arcsin (x/3) \) and \( dx = 3 cos u \; du \)
\( 9 - x^2 = 9 - 9 \sin^2 u \)
\( = 9(1-\sin^2) = 9 \cos^2 u \)
and
\( \sqrt{9-x^2} = \sqrt{9 \cos^2 u} = 3 \cos u \)
Substitute and write
\( \displaystyle I = \int \sqrt{9-x^2} \; dx = \int (3 \cos u ) \; 3 \cos u \; du \)
\( \displaystyle = 9 \int \cos^2 u du \)
Use power reduction trigonometric identity to write \( \cos^2 u = \dfrac{\cos(2u)+1}{2} \)
\( \displaystyle = \dfrac{9}{2} \int \left( \cos(2u)+1\right) du \)
Evaluate
\( \displaystyle I = \dfrac{9}{2} \left( \dfrac{\sin (2u)}{2} + u \right) \)
Use the trigonomrtic identity \( \sin (2u) = 2 \sin u \cos u \)
\( \displaystyle I = \dfrac{9}{2} \left( \sin u \cos u + u \right) \)
Substitute \( u \) by \( \arcsin (x/3) \) , \( \sin u = \dfrac{x}{3} \) , \( \sin u = \sqrt {1-sin^2 u } = \sqrt {1 - x^2/9} \) to rewrite
\( \displaystyle I = \dfrac{9}{2} \left( \dfrac{x}{3} \sqrt {1 - \dfrac{x^2}{9}} + \arcsin \left(\dfrac{x}{3}\right) \right) \)
\( \displaystyle I = \dfrac{x}{2} \sqrt {9 - x^2} + \dfrac{9}{2} \arcsin \left(\dfrac{x}{3} \right) \)
More Questions with Answers
Part 1
Find the volume of the solid bounded by the planes \( x = 1 \) , \( x =2 \) , \( y = 0 \) , \( y = 2 \) , \( z = 0 \) and lies under the hyperbolic paraboloid \( z = 20 + x^2 - 2y^2 \).
Find the volume of the solid bounded by the planes \( y = 2 \) , \( z = 0 \) and lies under the surface \( z = 9-x^2 \).
Find the volume of the solid bounded by the cylinder \( 9 - x^2 - y^2 = 0 \) , \( z = 0 \) and lies under the plane \( z=5+0.5x+y \).
Find area of the region in the \( x y \) plane enclosed by the circle of equation \( x^2 + y^2 = 4 \) and the line \( y = - 1 \).
Joel Hass, University of California, Davis; Maurice D. Weir Naval Postgraduate School; George B. Thomas, Jr.Massachusetts Institute of Technology ; University Calculus , Early Transcendentals, Third Edition
, Boston Columbus , 2016, Pearson.
