Calculus Examples

$r = 13 \cos (θ)$ , $r = 6 + \cos (θ)$

Step 1

Solve by substitution to find the intersection between the curves.

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Step 1.1

Eliminate the equal sides of each equation and combine.

$13 \cos (θ) = 6 + \cos (θ)$

Step 1.2

Solve $13 \cos (θ) = 6 + \cos (θ)$ for $θ$ .

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Step 1.2.1

Move all terms containing $\cos (θ)$ to the left side of the equation.

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Step 1.2.1.1

Subtract $\cos (θ)$ from both sides of the equation.

$13 \cos (θ) - \cos (θ) = 6$

Step 1.2.1.2

Subtract $\cos (θ)$ from $13 \cos (θ)$ .

$12 \cos (θ) = 6$

Step 1.2.2

Divide each term in $12 \cos (θ) = 6$ by $12$ and simplify.

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Step 1.2.2.1

Divide each term in $12 \cos (θ) = 6$ by $12$ .

$\frac{12 \cos (θ)}{12} = \frac{6}{12}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Cancel the common factor of $12$ .

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Step 1.2.2.2.1.1

Cancel the common factor.

$\frac{12 \cos (θ)}{12} = \frac{6}{12}$

Step 1.2.2.2.1.2

Divide $\cos (θ)$ by $1$ .

$\cos (θ) = \frac{6}{12}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Cancel the common factor of $6$ and $12$ .

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Step 1.2.2.3.1.1

Factor $6$ out of $6$ .

$\cos (θ) = \frac{6 (1)}{12}$

Step 1.2.2.3.1.2

Cancel the common factors.

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Step 1.2.2.3.1.2.1

Factor $6$ out of $12$ .

$\cos (θ) = \frac{6 \cdot 1}{6 \cdot 2}$

Step 1.2.2.3.1.2.2

Cancel the common factor.

$\cos (θ) = \frac{6 \cdot 1}{6 \cdot 2}$

Step 1.2.2.3.1.2.3

Rewrite the expression.

$\cos (θ) = \frac{1}{2}$

Step 1.2.3

Take the inverse cosine of both sides of the equation to extract $θ$ from inside the cosine.

$θ = \arccos (\frac{1}{2})$

Step 1.2.4

Simplify the right side.

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Step 1.2.4.1

The exact value of $\arccos (\frac{1}{2})$ is $\frac{π}{3}$ .

$θ = \frac{π}{3}$

Step 1.2.5

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$θ = 2 π - \frac{π}{3}$

Step 1.2.6

Simplify $2 π - \frac{π}{3}$ .

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Step 1.2.6.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$θ = 2 π \cdot \frac{3}{3} - \frac{π}{3}$

Step 1.2.6.2

Combine fractions.

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Step 1.2.6.2.1

Combine $2 π$ and $\frac{3}{3}$ .

$θ = \frac{2 π \cdot 3}{3} - \frac{π}{3}$

Step 1.2.6.2.2

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 3 - π}{3}$

Step 1.2.6.3

Simplify the numerator.

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Step 1.2.6.3.1

Multiply $3$ by $2$ .

$θ = \frac{6 π - π}{3}$

Step 1.2.6.3.2

Subtract $π$ from $6 π$ .

$θ = \frac{5 π}{3}$

Step 1.2.7

Find the period of $\cos (θ)$ .

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Step 1.2.7.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 1.2.7.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 1.2.7.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 1.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 1.2.8

The period of the $\cos (θ)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$θ = \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Step 1.3

Evaluate $r$ when $θ = \frac{π}{3} + 2 π n$ .

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Step 1.3.1

Substitute $\frac{π}{3} + 2 π n$ for $θ$ .

$r = 6 + \cos (\frac{π}{3} + 2 π n)$

Step 1.3.2

Substitute $\frac{π}{3} + 2 π n$ for $θ$ in $r = 6 + \cos (\frac{π}{3} + 2 π n)$ and solve for $r$ .

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Step 1.3.2.1

Remove parentheses.

$r = 6 + \cos (\frac{π}{3} + 2 π n)$

Step 1.3.2.2

Reorder $\frac{π}{3}$ and $2 π n$ .

$r = 6 + \cos (2 π n + \frac{π}{3})$

Step 1.4

Evaluate $r$ when $θ = \frac{5 π}{3} + 2 π n$ .

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Step 1.4.1

Substitute $\frac{5 π}{3} + 2 π n$ for $θ$ .

$r = 6 + \cos (\frac{5 π}{3} + 2 π n)$

Step 1.4.2

Substitute $\frac{5 π}{3} + 2 π n$ for $θ$ in $r = 6 + \cos (\frac{5 π}{3} + 2 π n)$ and solve for $r$ .

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Step 1.4.2.1

Remove parentheses.

$r = 6 + \cos (\frac{5 π}{3} + 2 π n)$

Step 1.4.2.2

Reorder $\frac{5 π}{3}$ and $2 π n$ .

$r = 6 + \cos (2 π n + \frac{5 π}{3})$

Step 1.5

The solution of the system of equations is all the values that make the system true.

$r = 6 + \cos (2 π n + \frac{π}{3}), θ = \frac{π}{3} + 2 π n r = 6 + \cos (2 π n + \frac{5 π}{3}), θ = \frac{5 π}{3} + 2 π n$

Step 1.6

List all of the solutions.

$r = 6 + \cos (2 π n + \frac{π}{3}), θ = \frac{π}{3} + 2 π n$

$r = 6 + \cos (2 π n + \frac{5 π}{3}), θ = \frac{5 π}{3} + 2 π n$

$r = 6 + \cos (2 π n + \frac{π}{3}), θ = \frac{π}{3} + 2 π n$

$r = 6 + \cos (2 π n + \frac{5 π}{3}), θ = \frac{5 π}{3} + 2 π n$

Step 2

The area between the given curves is unbounded.

Unbounded area

Step 3