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.
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 \( 2 \)
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.
Solution 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.
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.
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 \(C(a/6,a/3)\)
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.
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.
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.