Example 1
Question: Calculate the double integral \( \displaystyle V = \iint_R \sqrt {1 - x^2 - y^2} \;dy \;dx \) where the region \( R \) is the surface enclosed by a circle on the plane \( xy\)-plane with center at the origin and radius equal to \( 1\).
Solution to Example 1
The given integral is in rectangular coordinates and cannot be done using elementary functions. Let us try to convert to polar coordinates.
Let \( f(x,y) = \sqrt {1 - x^2 - y^2} \) and express it in polar coordinates.
\( r^2 = x^2 + y^2 \)
Hence function \( f(x,y) \) in polar form is given by
\( f(r,\theta) = \sqrt {1 - r^2} \)
Region \( R \) of integration is a circle and may be defined by inequalities as follows:
\( R: \) \( 0 \le \theta \le 2\pi \) and \( 0 \le r \le 1 \)
The given integral in rectangular coordinates may be converted to polar coordinates as follows
\( \displaystyle V = \iint_R \sqrt {1 - x^2 - y^2} \;dy \;dx = \int_0^{2\pi} \int_0^{1} \sqrt{1-r^2} \; r \; dr \; d\theta \)
Let \( I \) be the inner integral defined by
\( \displaystyle I = \int_0^{1} \sqrt{1-r^2} \; r \; dr \)
\( = \left[ -\frac{1}{3}\left(1-r^2\right)^{\frac{3}{2}} \right]_0^1 = \dfrac{1}{3} \)
Substitute \( I \) and calculate \( V \)
\( \displaystyle V = \int_0^{2\pi} (1/3) \; d\theta \)
\( = \dfrac{1}{3} \left[ \theta \right]_0^{2\pi} \)
\( = \dfrac{2\pi}{3} \)
Example 2
Question: Calculate the double integral \( \displaystyle V = \int_0^{1} \int_0^{\sqrt{1-x^2}} e^{x^2+y^2} \; dy \; dx \)
Solution to Example 2
The given integral cannot be easily calculated in rectangular coordinates hence the need to use polar coordinates instead which may make easy to evaluate.
Let us express \( f(x,y) = e^{x^2+y^2} \) in polar coordinates.
\( r^2 = x^2 + y^2 \)
Hence the function \( f(x,y) \) in polar form is given by
\( f(r,\theta) = e^{r^2} \)
In rectangular coordinates, the region \( R \) of integration is defined by the given limits of integration.
\( R: \) \( 0 \le x \le 1 \) and \( 0 \le y \le \sqrt{1-x^2} \)
Let us solve the inequality \( y \le \sqrt{1-x^2} \) graphically
Square both sides of the inequality
\( y^2 \le 1-x^2 \)
Rewrite with terms in \( x \) and \( y \) on the left side
\( y^2 + x^2 \le 1 \)
The above inequality is the set of all points \( (x,y) \) inside or on the circle with center at the origin \( (0,0) \) and radius \( 1 \)
Putting all inequalities together, the region of integration \( R \) is shown below.
The above region in polar form is shown below
and is defined by the inequalities
\( R: \) \( 0 \le \theta \le \pi/2 \) and \( 0 \le r \le 1 \)
The integral may now be converted in polar coordinates as follows
\( \displaystyle V = \int_0^{1} \int_0^{\sqrt{1-x^2}} e^{x^2+y^2} \; dy \; dx = \int_0^{\pi/2} \int_0^1 e^{r^2} r \; dr \; d\theta \)
Let \( I \) be the inner integral defined by
\( \displaystyle I = \int_0^1 e^{r^2} r \; dr \)
\( = \left[ \dfrac{1}{2} e{r^2} \right]_0^1 \)
\( = \dfrac{1}{2} (e - 1) \)
Substitute \( I \) into \( V \)
\( \displaystyle V = \int_0^{\pi/2} \dfrac{1}{2} (e - 1) \; d\theta \)
\( = \dfrac{\pi}{4} (e - 1) \)
Example 3
Question: Express the integral \( \displaystyle V = \iint_R {x^2+y^2} \; dy \; dx \) where \( R \) is a circle on the \( xy \) plane with center at the point \( (1,0) \) and a radius equal to \( 1 \) using Cartesian (or rectangular) and polar coordinates? Evaluate the integral.
Evaluate the integral.
Solution to Example 3
We first graph the region of integration \( R \) and define it in terms of rectangular and polar coordinates.
1) Rectangular coordinates
The equation of the circle with center at \( (1,0) \) and radius equal to \( 1 \) is given by
\( (x-1)^2 + y^2 = 1 \)
Expand and group like terms
\( x^2 - 2x + y^2 = 0 \)
Solve the above equation for y to obtain two solutions
\( y = \pm \sqrt {2x-x^2} \)
Using vertical strips, the region \( R \) may be described by the inequalities
\( R: \) \( 0 \le x \le 2 \) and \( - \sqrt {2x-x^2} \le y \le \sqrt {2x-x^2} \)
The integral may be written as
\( \displaystyle V = \int_0^2 \int_{-\sqrt {2x-x^2}}^{\sqrt {2x-x^2}} \; \; (x^2+y^2) \; dy \; dx \)
Let \( I_1 \) be the inner integral given by
\( \displaystyle I_1 = \int_{-\sqrt {2x-x^2}}^{\sqrt {2x-x^2}} \; \; (x^2+y^2) \; dy \)
Evaluate \( I_1 \)
\( \displaystyle I_1 = 2x^2\sqrt{2x-x^2}+2\cdot \frac{\left(2x-x^2\right)^{\frac{3}{2}}}{3} \)
The next step to calculate the outer integral above integral in rectangular coordinates is a challenging one.
2) Polar coordinates
Using strips from the origin to a point on the circle: at the origin \( r = 0 \); on the circle \( r = \sqrt {x^2 + y^2} \)
Let us convert the equation of the circle, \( x^2 + y^2 = 2x \) in polar form.
From relationship between rectangular and polar coordinates [6], we have
\( r^2 = x^2 + y^2 \) and \( x = r \cos \theta \)
Substitute \( x^2 + y^2 \) by \( r^2 \) and \( x \) by \( r \cos \theta \) in the equation \( x^2 + y^2 = 2x \) to obtain
\( r^2 = 2 r \cos \theta \)
Divide both sides of the above equation by \( r \)
\( r = 2 \cos \theta \)
In polar coordinates, the region \( R \) may be described by the inequalities
\( R: \) \( -\pi/2 \le \theta \le \pi/2 \) and \( 0 \le r \le 2 \cos \theta \)
Use \( r^2 = x^2 + y^2 \)
to write the integral in polar form as
\( \displaystyle V = \int_{-\pi/2}^{\pi/2} \int_{0}^{2 \cos \theta} r^2 \; r \; dr \; d\theta \)
Let the innner integral \(I \) be defined by
\( \displaystyle I = \int_{0}^{2 \cos \theta} r^2 \; r \; dr \)
Evaluate
\( \displaystyle I = \left[ r^4 / 4 \right]_{0}^{2 \cos \theta} \)
\( \displaystyle I = 4 cos^4 \theta \) (eq 1)
Reduce the power in the above expression \( \cos^4 \theta \)
\( \cos^4 \theta = \cos^2 \theta \cos^2 \theta \)
Use the power reducing identity
\( \cos^2 \theta = \dfrac{cos (2\theta) + 1}{2} \)
Hence
\( \cos^4 \theta = \left( \dfrac{\cos (2\theta) + 1}{2} \right)^2 \)
\( = \dfrac{\cos^2(2\theta) + 2 \cos (2\theta) + 1}{4} \)
Use the power reducing identity one more time to obtain
\( \cos^4 \theta = \dfrac{\cos(4\theta) + 4\cos(2\theta) }{8} + 3/8 \)
We now substitute the above expression in (eq 1) to obtain
\( I = \dfrac{\cos(4\theta) + 4\cos(2\theta) }{2} + 3/2 \)
Substitute \( I \) in the integral and evaluate
\( \displaystyle V = \int_{-\pi/2}^{\pi/2} (\dfrac{\cos(4\theta) + 4\cos(2\theta) }{2} + 3/2 ) \; d\theta \)
\( \displaystyle = \left[ \frac{1}{2}\left(\frac{1}{4}\sin \left(4\theta\right)+2\sin \left(2\theta\right)\right)+\frac{3}{2}\theta \right]_{-\pi/2}^{\pi/2} \)
\( = \dfrac{3\pi}{2} \)
The given integral is much easier evaluated using polar coordinates.
Example 4
Question: Change the integral \( \displaystyle V = \int_{-1}^0 \int_{-\sqrt{1-x^2}}^0 \dfrac{\sqrt{x^2+y^2}}{1+\sqrt{x^2+y^2}} \; dy \; dx \) into polar coordinates and evaluate it.
Solution to Example 4
From the limits of integration in rectangular coordinates, we deduce the region \( R \) of integration which is a quarter of a circle in quadrant III as
\( R: \) \( 0 \le x \le 2 \) and \( - \sqrt {1-x^2} \le y \le 0 \)
and its graph as shown below
In polar Polar coordinates using strips from the origin to a point on the quarter of a circle: at the origin \( r = 0 \). On the circle \( r = 1 \), the region \( R \) of integration in polar coordinates may be defined as
\( R: \) \( \pi \le \theta \le 3\pi/2 \) and \( 0 \le r \le 1 \)
Use \( r = \sqrt { x^2 + y^2 } \)
to write the given integral in polar form as
\( \displaystyle V = \int_{\pi}^{3\pi/2} \int_{0}^{1} \dfrac{r}{1+r} \; r \; dr \; d\theta \)
\( \displaystyle = \int_{\pi}^{3\pi/2} \int_{0}^{1} \dfrac{r^2}{1+r} \; dr \; d\theta \)
Use division to expand the integrand \( \dfrac{r^2}{1+r} \) as follows
\( \dfrac{r^2}{1+r} = r-1+\frac{1}{r+1} \)
Substitute the integrand in expanded for in the integral
\( \displaystyle V = \int_{\pi}^{3\pi/2} \int_{0}^{1} (r-1+\frac{1}{r+1}) \; dr \; d\theta \)
Integrate
\( \displaystyle = \int_{\pi}^{3\pi/2} \left[ \dfrac{r^2}{2} - r + ln |r+1| \right]_0^1 d\theta \)
Evaluate
\( \displaystyle = \int_{\pi}^{3\pi/2} (\ln (2)-\dfrac{1}{2}) d\theta \)
Integrate
\( = \left( \ln (2)-\dfrac{1}{2} \right) \left[ \; \theta \; \right]_{\pi}^{3\pi/2} \)
Evaluate
\( V = \dfrac{\pi}{2} \left( \ln (2)-\dfrac{1}{2} \right) \)