(57) Finding Area In Polar Coordinates

Finding Area Enclosed by Polar Curves

Example 1: Polar Equation ( r = \sin(2\theta) )

• Graph the Curve

  • The curve consists of four petals, and we will focus on finding the area of one petal.

Formula for Area

• Use the formula:[ A = \int_{\alpha}^{\beta} \frac{1}{2} r^2 , d\theta ]

• For ( r = \sin(2\theta) ), we find ( \alpha ) and ( \beta ):

  • The petal is bounded by ( \theta = 0 ) to ( \theta = \frac{\pi}{2} ).

Setting Up the Integral

• Area becomes:[ A = \int_{0}^{\frac{\pi}{2}} \frac{1}{2} \sin^2(2\theta) , d\theta ]

Simplifying the Integral

• Use the power-reducing formula:

  • ( \sin^2(\theta) = \frac{1}{2}(1 - \cos(2\theta)) )

  • Thus, ( \sin^2(2\theta) = \frac{1}{2}(1 - \cos(4\theta)) )

• Now the integral can be expressed as:[ A = \int_{0}^{\frac{\pi}{2}} \frac{1}{4}(1 - \cos(4\theta)) , d\theta ]

Evaluating the Integral

• Calculate the antiderivative:

  • Antiderivative of ( 1 ) is ( \theta ) and of ( \cos(4\theta) ) is ( \frac{\sin(4\theta)}{4} ).

• Evaluate from ( \theta = 0 ) to ( \frac{\pi}{2} ):

  • At ( \theta = \frac{\pi}{2} ), ( \sin(2\pi) = 0 ); at ( \theta = 0 ), ( \sin(0) = 0 ).

• Thus the area of one petal is:[ A = \frac{1}{4} \times \frac{\pi}{2} = \frac{\pi}{8} ]

Example 2: Polar Equation ( r = 2\cos(3\theta) )

• This graph has three petals.

Finding Area of One Petal

• Use the area formula:[ A = \int_{\alpha}^{\beta} \frac{1}{2} r^2 , d\theta ]

• Here, find angles ( \alpha ) and ( \beta ):

  • First petal is bounded by ( -\frac{\pi}{6} ) and ( \frac{\pi}{6} ).

• Area becomes:[ A = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{1}{2} (2\cos(3\theta))^2 , d\theta ]

Simplifying and Evaluating the Integral

• This leads to:[ 2 \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \cos^2(3\theta) , d\theta ]

• Using the power-reducing formula gives: [ \cos^2(3\theta) = \frac{1}{2}(1 + \cos(6\theta)) ]

• Evaluate the integral, ultimately leading to:[ A = \frac{\pi}{3} ] for one petal.

• For three petals, multiply by 3, total area = ( \pi ).

Example 3: Polar Equation ( r^2 = 4\cos(2\theta) )

• This corresponds to a lemniscate shape.

Finding Total Area

• Use similar approach to find the area of one loop and multiply by 2 for the total area.

• First loop is bounded by ( -\frac{\pi}{4} ) to ( \frac{\pi}{4} ):[ A = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 2 , \cos(2\theta) , d\theta ]

Evaluation

• The result will give:

  • Area of one loop = ( 2 ) and total area = ( 4 ).

Example 4: Polar Equation ( r = 3\cos(\theta) )

• This is a circle with radius ( \frac{3}{2} ).

Area Calculation

• Use area formula from ( -\frac{\pi}{2} ) to ( \frac{\pi}{2} ):[ A = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{9}{2} , d\theta ]

• Result gives:[ A = \frac{9\pi}{4} ]

Example 5: Polar Equation ( r = 1 + 2\sin(\theta) )

• Focus on finding the area of the inner loop.

Identifying Boundaries

• Use ( r = 0 ) to find angles:

  • Set ( 1 + 2\sin(\theta) = 0 ) leading to ( a = \frac{7\pi}{6}, b = \frac{11\pi}{6} ).

• Setup area integral:[ A = \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} \frac{1}{2} (1 + 2\sin(\theta))^2 , d\theta ]

General Formula

• Whenever finding area of polar curves, use:[ A = \int_{\alpha}^{\beta} \frac{1}{2} r^2 , d\theta ]

• Identify bounds and square the function accordingly.