Calculus Examples

$z = 4 - y^{2}$ , $x^{2} + y^{2} = 2 x$ , $z = 0$

Step 1

Solve by substitution to find the intersection between the curves.

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Step 1.1

Replace all occurrences of $z$ with $0$ in each equation.

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Step 1.1.1

Replace all occurrences of $z$ in $z = 4 - y^{2}$ with $0$ .

$0 = 4 - y^{2}$

$z = 0$

$x^{2} + y^{2} = 2 x$

Step 1.1.2

Simplify the left side.

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Step 1.1.2.1

Remove parentheses.

$0 = 4 - y^{2}$

$z = 0$

$x^{2} + y^{2} = 2 x$

$0 = 4 - y^{2}$

$z = 0$

$x^{2} + y^{2} = 2 x$

$0 = 4 - y^{2}$

$z = 0$

$x^{2} + y^{2} = 2 x$

Step 1.2

Solve for $y$ in $0 = 4 - y^{2}$ .

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Step 1.2.1

Rewrite the equation as $4 - y^{2} = 0$ .

$4 - y^{2} = 0$

$z = 0$

$x^{2} + y^{2} = 2 x$

Step 1.2.2

Subtract $4$ from both sides of the equation.

$- y^{2} = - 4$

$z = 0$

$x^{2} + y^{2} = 2 x$

Step 1.2.3

Divide each term in $- y^{2} = - 4$ by $- 1$ and simplify.

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Step 1.2.3.1

Divide each term in $- y^{2} = - 4$ by $- 1$ .

$\frac{- y^{2}}{- 1} = \frac{- 4}{- 1}$

$z = 0$

$x^{2} + y^{2} = 2 x$

Step 1.2.3.2

Simplify the left side.

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Step 1.2.3.2.1

Dividing two negative values results in a positive value.

$\frac{y^{2}}{1} = \frac{- 4}{- 1}$

$z = 0$

$x^{2} + y^{2} = 2 x$

Step 1.2.3.2.2

Divide $y^{2}$ by $1$ .

$y^{2} = \frac{- 4}{- 1}$

$z = 0$

$x^{2} + y^{2} = 2 x$

$y^{2} = \frac{- 4}{- 1}$

$z = 0$

$x^{2} + y^{2} = 2 x$

Step 1.2.3.3

Simplify the right side.

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Step 1.2.3.3.1

Divide $- 4$ by $- 1$ .

$y^{2} = 4$

$z = 0$

$x^{2} + y^{2} = 2 x$

$y^{2} = 4$

$z = 0$

$x^{2} + y^{2} = 2 x$

$y^{2} = 4$

$z = 0$

$x^{2} + y^{2} = 2 x$

Step 1.2.4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{4}$

$z = 0$

$x^{2} + y^{2} = 2 x$

Step 1.2.5

Simplify $\pm \sqrt{4}$ .

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Step 1.2.5.1

Rewrite $4$ as $2^{2}$ .

$y = \pm \sqrt{2^{2}}$

$z = 0$

$x^{2} + y^{2} = 2 x$

Step 1.2.5.2

Pull terms out from under the radical, assuming positive real numbers.

$y = \pm 2$

$z = 0$

$x^{2} + y^{2} = 2 x$

$y = \pm 2$

$z = 0$

$x^{2} + y^{2} = 2 x$

Step 1.2.6

The complete solution is the result of both the positive and negative portions of the solution.

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Step 1.2.6.1

First, use the positive value of the $\pm$ to find the first solution.

$y = 2$

$z = 0$

$x^{2} + y^{2} = 2 x$

Step 1.2.6.2

Next, use the negative value of the $\pm$ to find the second solution.

$y = - 2$

$z = 0$

$x^{2} + y^{2} = 2 x$

Step 1.2.6.3

The complete solution is the result of both the positive and negative portions of the solution.

$y = 2, - 2$

$z = 0$

$x^{2} + y^{2} = 2 x$

$y = 2, - 2$

$z = 0$

$x^{2} + y^{2} = 2 x$

$y = 2, - 2$

$z = 0$

$x^{2} + y^{2} = 2 x$

Step 1.3

Solve the system $y = 2, z = 0, x^{2} + y^{2} = 2 x$ .

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Step 1.3.1

Replace all occurrences of $y$ with $2$ in each equation.

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Step 1.3.1.1

Replace all occurrences of $y$ in $x^{2} + y^{2} = 2 x$ with $2$ .

$x^{2} + {(2)}^{2} = 2 x$

$y = 2$

$z = 0$

Step 1.3.1.2

Simplify the left side.

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Step 1.3.1.2.1

Raise $2$ to the power of $2$ .

$x^{2} + 4 = 2 x$

$y = 2$

$z = 0$

$x^{2} + 4 = 2 x$

$y = 2$

$z = 0$

$x^{2} + 4 = 2 x$

$y = 2$

$z = 0$

Step 1.3.2

Solve for $x$ in $x^{2} + 4 = 2 x$ .

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Step 1.3.2.1

Subtract $2 x$ from both sides of the equation.

$x^{2} + 4 - 2 x = 0$

$y = 2$

$z = 0$

Step 1.3.2.2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

$y = 2$

$z = 0$

Step 1.3.2.3

Substitute the values $a = 1$ , $b = - 2$ , and $c = 4$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot 4)}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.4

Simplify.

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Step 1.3.2.4.1

Simplify the numerator.

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Step 1.3.2.4.1.1

Raise $- 2$ to the power of $2$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.4.1.2

Multiply $- 4 \cdot 1 \cdot 4$ .

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Step 1.3.2.4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 4}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.4.1.2.2

Multiply $- 4$ by $4$ .

$x = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

$y = 2$

$z = 0$

$x = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.4.1.3

Subtract $16$ from $4$ .

$x = \frac{2 \pm \sqrt{- 12}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.4.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.4.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.4.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 1.3.2.4.1.7.1

Factor $4$ out of $12$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.4.1.7.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

$y = 2$

$z = 0$

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.4.1.8

Pull terms out from under the radical.

$x = \frac{2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.4.1.9

Move $2$ to the left of $i$ .

$x = \frac{2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

$y = 2$

$z = 0$

$x = \frac{2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.4.2

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 i \sqrt{3}}{2}$

$y = 2$

$z = 0$

Step 1.3.2.4.3

Simplify $\frac{2 \pm 2 i \sqrt{3}}{2}$ .

$x = 1 \pm i \sqrt{3}$

$y = 2$

$z = 0$

$x = 1 \pm i \sqrt{3}$

$y = 2$

$z = 0$

Step 1.3.2.5

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 1.3.2.5.1

Simplify the numerator.

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Step 1.3.2.5.1.1

Raise $- 2$ to the power of $2$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.5.1.2

Multiply $- 4 \cdot 1 \cdot 4$ .

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Step 1.3.2.5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 4}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.5.1.2.2

Multiply $- 4$ by $4$ .

$x = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

$y = 2$

$z = 0$

$x = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.5.1.3

Subtract $16$ from $4$ .

$x = \frac{2 \pm \sqrt{- 12}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.5.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.5.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.5.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.5.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 1.3.2.5.1.7.1

Factor $4$ out of $12$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.5.1.7.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

$y = 2$

$z = 0$

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.5.1.8

Pull terms out from under the radical.

$x = \frac{2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.5.1.9

Move $2$ to the left of $i$ .

$x = \frac{2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

$y = 2$

$z = 0$

$x = \frac{2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.5.2

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 i \sqrt{3}}{2}$

$y = 2$

$z = 0$

Step 1.3.2.5.3

Simplify $\frac{2 \pm 2 i \sqrt{3}}{2}$ .

$x = 1 \pm i \sqrt{3}$

$y = 2$

$z = 0$

Step 1.3.2.5.4

Change the $\pm$ to $+$ .

$x = 1 + i \sqrt{3}$

$y = 2$

$z = 0$

$x = 1 + i \sqrt{3}$

$y = 2$

$z = 0$

Step 1.3.2.6

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 1.3.2.6.1

Simplify the numerator.

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Step 1.3.2.6.1.1

Raise $- 2$ to the power of $2$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.6.1.2

Multiply $- 4 \cdot 1 \cdot 4$ .

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Step 1.3.2.6.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 4}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.6.1.2.2

Multiply $- 4$ by $4$ .

$x = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

$y = 2$

$z = 0$

$x = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.6.1.3

Subtract $16$ from $4$ .

$x = \frac{2 \pm \sqrt{- 12}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.6.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.6.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.6.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.6.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 1.3.2.6.1.7.1

Factor $4$ out of $12$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.6.1.7.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

$y = 2$

$z = 0$

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.6.1.8

Pull terms out from under the radical.

$x = \frac{2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.6.1.9

Move $2$ to the left of $i$ .

$x = \frac{2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

$y = 2$

$z = 0$

$x = \frac{2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

$y = 2$

$z = 0$

Step 1.3.2.6.2

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 i \sqrt{3}}{2}$

$y = 2$

$z = 0$

Step 1.3.2.6.3

Simplify $\frac{2 \pm 2 i \sqrt{3}}{2}$ .

$x = 1 \pm i \sqrt{3}$

$y = 2$

$z = 0$

Step 1.3.2.6.4

Change the $\pm$ to $-$ .

$x = 1 - i \sqrt{3}$

$y = 2$

$z = 0$

$x = 1 - i \sqrt{3}$

$y = 2$

$z = 0$

Step 1.3.2.7

The final answer is the combination of both solutions.

$x = 1 + i \sqrt{3}, 1 - i \sqrt{3}$

$y = 2$

$z = 0$

$x = 1 + i \sqrt{3}, 1 - i \sqrt{3}$

$y = 2$

$z = 0$

Step 1.3.3

Solve the system of equations.

$x = 1 + i \sqrt{3}, y = 2, z = 0$

Step 1.3.4

Solve the system of equations.

$x = 1 - i \sqrt{3}, y = 2, z = 0$

Step 1.4

Solve the system $y = - 2, z = 0, x^{2} + y^{2} = 2 x$ .

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Step 1.4.1

Replace all occurrences of $y$ with $- 2$ in each equation.

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Step 1.4.1.1

Replace all occurrences of $y$ in $x^{2} + y^{2} = 2 x$ with $- 2$ .

$x^{2} + {(- 2)}^{2} = 2 x$

$y = - 2$

$z = 0$

Step 1.4.1.2

Simplify the left side.

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Step 1.4.1.2.1

Raise $- 2$ to the power of $2$ .

$x^{2} + 4 = 2 x$

$y = - 2$

$z = 0$

$x^{2} + 4 = 2 x$

$y = - 2$

$z = 0$

$x^{2} + 4 = 2 x$

$y = - 2$

$z = 0$

Step 1.4.2

Solve for $x$ in $x^{2} + 4 = 2 x$ .

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Step 1.4.2.1

Subtract $2 x$ from both sides of the equation.

$x^{2} + 4 - 2 x = 0$

$y = - 2$

$z = 0$

Step 1.4.2.2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

$y = - 2$

$z = 0$

Step 1.4.2.3

Substitute the values $a = 1$ , $b = - 2$ , and $c = 4$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot 4)}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.4

Simplify.

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Step 1.4.2.4.1

Simplify the numerator.

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Step 1.4.2.4.1.1

Raise $- 2$ to the power of $2$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.4.1.2

Multiply $- 4 \cdot 1 \cdot 4$ .

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Step 1.4.2.4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 4}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.4.1.2.2

Multiply $- 4$ by $4$ .

$x = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

$y = - 2$

$z = 0$

$x = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.4.1.3

Subtract $16$ from $4$ .

$x = \frac{2 \pm \sqrt{- 12}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.4.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.4.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.4.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 1.4.2.4.1.7.1

Factor $4$ out of $12$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.4.1.7.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

$y = - 2$

$z = 0$

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.4.1.8

Pull terms out from under the radical.

$x = \frac{2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.4.1.9

Move $2$ to the left of $i$ .

$x = \frac{2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

$y = - 2$

$z = 0$

$x = \frac{2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.4.2

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 i \sqrt{3}}{2}$

$y = - 2$

$z = 0$

Step 1.4.2.4.3

Simplify $\frac{2 \pm 2 i \sqrt{3}}{2}$ .

$x = 1 \pm i \sqrt{3}$

$y = - 2$

$z = 0$

$x = 1 \pm i \sqrt{3}$

$y = - 2$

$z = 0$

Step 1.4.2.5

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 1.4.2.5.1

Simplify the numerator.

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Step 1.4.2.5.1.1

Raise $- 2$ to the power of $2$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.5.1.2

Multiply $- 4 \cdot 1 \cdot 4$ .

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Step 1.4.2.5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 4}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.5.1.2.2

Multiply $- 4$ by $4$ .

$x = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

$y = - 2$

$z = 0$

$x = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.5.1.3

Subtract $16$ from $4$ .

$x = \frac{2 \pm \sqrt{- 12}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.5.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.5.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.5.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.5.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 1.4.2.5.1.7.1

Factor $4$ out of $12$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.5.1.7.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

$y = - 2$

$z = 0$

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.5.1.8

Pull terms out from under the radical.

$x = \frac{2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.5.1.9

Move $2$ to the left of $i$ .

$x = \frac{2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

$y = - 2$

$z = 0$

$x = \frac{2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.5.2

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 i \sqrt{3}}{2}$

$y = - 2$

$z = 0$

Step 1.4.2.5.3

Simplify $\frac{2 \pm 2 i \sqrt{3}}{2}$ .

$x = 1 \pm i \sqrt{3}$

$y = - 2$

$z = 0$

Step 1.4.2.5.4

Change the $\pm$ to $+$ .

$x = 1 + i \sqrt{3}$

$y = - 2$

$z = 0$

$x = 1 + i \sqrt{3}$

$y = - 2$

$z = 0$

Step 1.4.2.6

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 1.4.2.6.1

Simplify the numerator.

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Step 1.4.2.6.1.1

Raise $- 2$ to the power of $2$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.6.1.2

Multiply $- 4 \cdot 1 \cdot 4$ .

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Step 1.4.2.6.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 4}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.6.1.2.2

Multiply $- 4$ by $4$ .

$x = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

$y = - 2$

$z = 0$

$x = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.6.1.3

Subtract $16$ from $4$ .

$x = \frac{2 \pm \sqrt{- 12}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.6.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.6.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.6.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.6.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 1.4.2.6.1.7.1

Factor $4$ out of $12$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.6.1.7.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

$y = - 2$

$z = 0$

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.6.1.8

Pull terms out from under the radical.

$x = \frac{2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.6.1.9

Move $2$ to the left of $i$ .

$x = \frac{2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

$y = - 2$

$z = 0$

$x = \frac{2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

$y = - 2$

$z = 0$

Step 1.4.2.6.2

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 i \sqrt{3}}{2}$

$y = - 2$

$z = 0$

Step 1.4.2.6.3

Simplify $\frac{2 \pm 2 i \sqrt{3}}{2}$ .

$x = 1 \pm i \sqrt{3}$

$y = - 2$

$z = 0$

Step 1.4.2.6.4

Change the $\pm$ to $-$ .

$x = 1 - i \sqrt{3}$

$y = - 2$

$z = 0$

$x = 1 - i \sqrt{3}$

$y = - 2$

$z = 0$

Step 1.4.2.7

The final answer is the combination of both solutions.

$x = 1 + i \sqrt{3}, 1 - i \sqrt{3}$

$y = - 2$

$z = 0$

$x = 1 + i \sqrt{3}, 1 - i \sqrt{3}$

$y = - 2$

$z = 0$

Step 1.4.3

Solve the system of equations.

$x = 1 + i \sqrt{3}, y = - 2, z = 0$

Step 1.4.4

Solve the system of equations.

$x = 1 - i \sqrt{3}, y = - 2, z = 0$

Step 1.5

List all of the solutions.

$x = 1 + i \sqrt{3}, y = 2, z = 0$

$x = 1 - i \sqrt{3}, y = 2, z = 0$

$x = 1 + i \sqrt{3}, y = - 2, z = 0$

$x = 1 - i \sqrt{3}, y = - 2, z = 0$

$x = 1 + i \sqrt{3}, y = 2, z = 0$

$x = 1 - i \sqrt{3}, y = 2, z = 0$

$x = 1 + i \sqrt{3}, y = - 2, z = 0$

$x = 1 - i \sqrt{3}, y = - 2, z = 0$

Step 2

The area between the given curves is unbounded.

Unbounded area

Step 3