What is the area bounded by the parametric equations? : # x=acos theta # and # y=bsin theta #

Question

Charles Arocha · Accepted Answer

It is vital when dealing with parametric equations (or polar coordinates) to get a full understanding of the effect of the parameter on the curve (and sign) so that positive and negative areas can be determined and dealt with.

We have parametric equations:

# x=acos theta #
# y=bsin theta #

Where #a,b gt 0#. Let us first look at how # x=acos theta # and # y=bsin theta # behave over the given domain #0 le theta le 2pi#

#0 le theta le pi/2 => x>0, y>0#
#pi/2 le theta le pi => x<0, y>0#
#pi le theta le (3pi)/2 => x<0, y<0#
#(3pi)/2 le theta le 2pi => x>0, y<0#

So we can see that as #theta# varies on the domain we move uniformly from Q1, Q2, Q3 then Q4. Thus #0 le theta le pi# (Q1,Q2) will contribute positively and #pi le theta le 2pi# (Q3,Q4) will contribute negatively.

If we now examine the parametric curve

We can see that we can evaluate the area by symmetry as #4# times that of Q1

Thus we can represent the area of the ellipse by:

# A = 4 \ int_(x=0)^(x=a) \ y \ dx #

Now we will change variable from #x# to #theta# to actually perform the integration:

# x=acos theta => dx/(d theta)=-asin theta #
# y=bsin theta #

And for the limits of integration we have:

# x=0 => acos theta =0 => theta=pi/2 #
# x=a => acos theta =a => theta=0 #

And so we can evaluate the integral as follows;

# A = 4 \ int_(pi/2)^(0) \ (bsin theta)(-asin theta) \ d theta #
# \ \ = -4ab \ int_(pi/2)^(0) \ sin^2 theta \ d theta #
# \ \ = 4ab \ int_(0)^(pi/2) \ sin^2 theta \ d theta #
# \ \ = 4ab \ int_(0)^(pi/2) \ sin^2 theta \ d theta #
# \ \ = 4ab \ int_(0)^(pi/2) \ (1-cos2theta)/2 \ d theta #
# \ \ = 2ab \ int_(0)^(pi/2) \ 1-cos2theta \ d theta #
# \ \ = 2ab \ [theta-(sin2theta)/2]_(0)^(pi/2)#
# \ \ = 2ab \ {(pi/2-0)-(0-0)}#
# \ \ = pi ab#

Paisley Johnson · Answer

undefined To find the area bounded by the parametric equations $ x = a \cos(\theta) $ and $ y = b \sin(\theta) $, where $ a $ and $ b $ are constants, you can use the formula for the area enclosed by a parametric curve:

$$ A = \frac{1}{2} \int_{\theta_1}^{\theta_2} (x \cdot y') \, d\theta $$

First, differentiate $ y = b \sin(\theta) $ with respect to $ \theta $ to find $ y' $. Then substitute $ x = a \cos(\theta) $ and $ y = b \sin(\theta) $ into the formula and integrate over the appropriate interval $ \theta_1 $ to $ \theta_2 $. This will give you the area enclosed by the parametric equations.