Given an ellipse with major and minor axis of #a# and #b# find the area of the ellipse? Show your work. the perimeter of the ellipse is rather difficult to derive in closed form, discuss how you would approach it?

Question

Sadie Adams · Accepted Answer

Part 2. Perimeter of an ellipse.

As for area part of the question let us consider an ellipse having center at origin #(0,0)# with semi-major axis and semi-minor axis of lengths #a and b# along #x#-axis and #y#-axis respectively as shown in the figure above.

The equation of this ellipse is written in the standard form

#x^2 / a^2 + y^2 / b^2 = 1# .......(1)
where it is assumed that #0 " < " b " < "a#

Now the ellipse can be represented by the parametric equations

#x=acostheta andy=bsintheta#
#0≤θ≤2π#

The ellipse is symmetrical with respect to #x and y# axes, therefore, the total perimeter #p# is #4times# the perimeter of one quadrant. Considering first quadrant.

Length #dp# of an infinitesimal element of ellipse #dx# at an angle #theta# from the #x#-axis is given by

#dp=sqrt(((dx)/(d theta))^2+((dy)/(d theta))^2)#
#:.p= 4int_0^(π/2)sqrt(((dx)/(d theta))^2+((dy)/(d theta))^2)cdot d theta#
#=>p= 4int_0^(π/2)sqrt(a^2sin^2theta+b^2cos^2theta)cdot d theta#
#=>p= 4int_0^(π/2)sqrt(a^2(1-cos^2theta)+b^2cos^2theta)cdot d theta#
#=>p= 4int_0^(π/2)sqrt(a^2-(a^2-b^2)cos^2theta)cdot d theta#

Multiplying and dividing with #a# we get

#=>p= 4int_0^(π/2)asqrt(1-(1-b^2/a^2)cos^2theta)cdot d theta#

Let us define eccentricity of ellipse #epsilon-=sqrt(1-b^2/a^2)#. Substituting in expression for #p# above we get

#p= 4aint_0^(π/2)sqrt(1-epsilon^2cos^2theta)cdot d theta#

This is called an elliptic integral and unfortunately can’t be evaluated using standard functions.

We can make use of Binomial theorem to rewrite the integrand as a sum of infinite series. Note that this series converges for all values of #theta# as #0<=epsilon^2cos^2theta" < "0#. Integrating term-wise the series we can find the perimeter of the ellipse.

Explicitly we get the following expression
#p=2pia[1-(1/2)^2epsilon^2/1-((1cdot3)/(2cdot4))^2epsilon^4/3-((1cdot3cdot5)/(2cdot4cdot6))^2epsilon^6/5....]#
.-.-.-.-.-.-.-.-.-.-.-

We may also note that for a circle of radius #a#, eccentricity #epsilon=0#. The expression for perimeter (circumference) reduces to

#p=2pia#

James Atkins · Answer

#1/4 piab#

Let us consider an ellipse having center at origin #(0,0)# with semi-major axis and semi-minor axis of lengths #a and b# along #x#-axis and #y#-axis respectively as shown in the figure above.

The equation of this ellipse can be written in the standard form

#x^2 / a^2 + y^2 / b^2 = 1# .......(1)
As the ellipse is symmetrical with respect to #x and y# axes, the total area #A# is #4times# the area in one quadrant.

Consider first quadrant only. Solving equation (1) for #y# we get

#y = b sqrt(1 - x^2 / a^2 )#
retaining only #+ve# root for the quadrant of interest.
Area #dA# of an infinitesimal vertical strip of ellipse of width #dx# at distance #x# is given by

#dA=b sqrt(( 1 - x^2 / a^2 ))cdot dx#

Area #A/4# is found by integrating this with respect to #x# over the limits #x=0 " to " x=a#

#A/4=int_0^ab sqrt(( 1 - x^2 / a^2 ))cdot dx# ......(2)
Substituting in (2) #x/a=sin t # and #dx = a cos t cdot dt#, with limits are from #t= 0" to "pi/2#, we get

#A/4=int_0^(pi/2) b sqrt(( 1-sin^2t ))cdotacostcdot dt#
Using #sqrt(( 1 - sin^2 t )) = cos t#

#=>A/4=abint_0^(pi/2) cos^2tcdot dt#

Rewriting using the identity #cos^2 t = ( cos 2t + 1 ) / 2#

#=>A/4=abint_0^(pi/2) ( cos 2t + 1 ) / 2cdot dt#

Evaluating the integral

#A/4 = 1/2 ab [ (sin 2t)/2 + t ]_0^( pi/2)#
#=>A/4 = 1/2 ab [ (sin pi)/2 + pi/2 ]#
#=>A/4 = (ab pi)/4#

#:.#Area of ellipse #A= pi a b ## =pixx"length of semi-major axis"xx"length of semi-minor axis"#
#color (white)( WWWWWWWWWWWWWWWWWWWWWWW)# ......(3)

In the given problem we have an ellipse with major axis and minor axis of lengths #a# and #b# respectively as shown below.

Inserting given values in equation (3) we get

#A=pixxa/2xxb/2#
#=>A=1/4 piab#

Perimeter part follows.

Sadie Adams · Answer

undefined To find the area of an ellipse with major axis $ a $ and minor axis $ b $, you can use the formula:

$$ \text{Area} = \pi \times a \times b $$

For deriving the perimeter of an ellipse in closed form, one common approach is to use elliptic integrals. Specifically, the perimeter of an ellipse cannot be expressed simply in terms of elementary functions like polynomials, exponentials, trigonometric functions, etc.

One method to approximate the perimeter is to use numerical integration techniques or series expansions. Another approach is to express the perimeter in terms of elliptic integrals, which are special functions that arise when dealing with ellipses. Elliptic integrals can be computed numerically, but they do not have a simple closed form expression.

So, to find the perimeter of an ellipse, one would typically resort to numerical methods or rely on approximate formulas derived from elliptic integrals.

Aaliyah Anderson · Answer

undefined To find the area $ A $ of an ellipse with major axis $ a $ and minor axis $ b $, we use the formula:

$$ A = \pi ab $$

To derive the perimeter of an ellipse in closed form, we can use approximations or numerical methods. One approach is to use elliptic integrals or series expansions to approximate the perimeter. Another method is to use numerical integration techniques to compute the arc length of the ellipse's boundary curve. These methods provide increasingly accurate approximations as the number of terms or the precision of numerical integration increases.