Find the Area Between the Curves y=16-x^2 , y=0 , x=-3 , x=2

$y = 16 - x^{2}$ , $y = 0$ , $x = - 3$ , $x = 2$

Solve by substitution to find the intersection between the curves.

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Substitute $0$ for $y$ into $y = 16 - x^{2}$ then solve for $x$ .

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Replace $y$ with $0$ in the equation.

$0 = 16 - x^{2}$

Solve the equation for $x$ .

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Rewrite the equation as $16 - x^{2} = 0$ .

$16 - x^{2} = 0$

Subtract $16$ from both sides of the equation.

$- x^{2} = - 16$

Multiply each term in $- x^{2} = - 16$ by $- 1$

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Multiply each term in $- x^{2} = - 16$ by $- 1$ .

$(- x^{2}) \cdot - 1 = (- 16) \cdot - 1$

Multiply $(- x^{2}) \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$1 x^{2} = (- 16) \cdot - 1$

Multiply $x^{2}$ by $1$ .

$x^{2} = (- 16) \cdot - 1$

Multiply $- 16$ by $- 1$ .

$x^{2} = 16$

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

$x = \pm \sqrt{16}$

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

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Simplify the right side of the equation.

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Rewrite $16$ as $4^{2}$ .

$x = \pm \sqrt{4^{2}}$

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

$x = \pm 4$

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

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

$x = 4$

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

$x = - 4$

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

$x = 4, - 4$

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(4, 0)$

$(- 4, 0)$

$(4, 0)$

$(- 4, 0)$

The area of the region between the curves is defined as the integral of the upper curve minus the integral of the lower curve over each region. The regions are determined by the intersection points of the curves. This can be done algebraically or graphically.

$A r e a = \int_{- 3}^{2} 16 - x^{2} d x - \int_{- 3}^{2} 0 d x$

Integrate to find the area between $- 3$ and $2$ .

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Combine the integrals into a single integral.

$\int_{- 3}^{2} 16 - x^{2} - (0) d x$

Subtract $0$ from $16 - x^{2}$ .

$\int_{- 3}^{2} 16 - x^{2} d x$

Split the single integral into multiple integrals.

$\int_{- 3}^{2} 16 d x + \int_{- 3}^{2} - x^{2} d x$

Since $16$ is constant with respect to $x$ , move $16$ out of the integral.

$16 x]_{- 3}^{2} + \int_{- 3}^{2} - x^{2} d x$

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

$16 x]_{- 3}^{2} - \int_{- 3}^{2} x^{2} d x$

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

$16 x]_{- 3}^{2} - (\frac{1}{3} x^{3}]_{- 3}^{2})$

Simplify the answer.

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Combine $\frac{1}{3}$ and $x^{3}$ .

$16 x]_{- 3}^{2} - (\frac{x^{3}}{3}]_{- 3}^{2})$

Substitute and simplify.

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Evaluate $16 x$ at $2$ and at $- 3$ .

$(16 \cdot 2) - 16 \cdot - 3 - (\frac{x^{3}}{3}]_{- 3}^{2})$

Evaluate $\frac{x^{3}}{3}$ at $2$ and at $- 3$ .

$16 \cdot 2 - 16 \cdot - 3 - (\frac{2^{3}}{3} - \frac{{(- 3)}^{3}}{3})$

Simplify.

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Multiply $16$ by $2$ .

$32 - 16 \cdot - 3 - (\frac{2^{3}}{3} - \frac{{(- 3)}^{3}}{3})$

Multiply $- 16$ by $- 3$ .

$32 + 48 - (\frac{2^{3}}{3} - \frac{{(- 3)}^{3}}{3})$

Add $32$ and $48$ .

$80 - (\frac{2^{3}}{3} - \frac{{(- 3)}^{3}}{3})$

Raise $2$ to the power of $3$ .

$80 - (\frac{8}{3} - \frac{{(- 3)}^{3}}{3})$

Raise $- 3$ to the power of $3$ .

$80 - (\frac{8}{3} - \frac{- 27}{3})$

Cancel the common factor of $- 27$ and $3$ .

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Factor $3$ out of $- 27$ .

$80 - (\frac{8}{3} - \frac{3 \cdot - 9}{3})$

Cancel the common factors.

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Factor $3$ out of $3$ .

$80 - (\frac{8}{3} - \frac{3 \cdot - 9}{3 (1)})$

Cancel the common factor.

$80 - (\frac{8}{3} - \frac{3 \cdot - 9}{3 \cdot 1})$

Rewrite the expression.

$80 - (\frac{8}{3} - \frac{- 9}{1})$

Divide $- 9$ by $1$ .

$80 - (\frac{8}{3} - - 9)$

Multiply $- 1$ by $- 9$ .

$80 - (\frac{8}{3} + 9)$

To write $9$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$80 - (\frac{8}{3} + 9 \cdot \frac{3}{3})$

Combine $9$ and $\frac{3}{3}$ .

$80 - (\frac{8}{3} + \frac{9 \cdot 3}{3})$

Combine the numerators over the common denominator.

$80 - \frac{8 + 9 \cdot 3}{3}$

Simplify the numerator.

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Multiply $9$ by $3$ .

$80 - \frac{8 + 27}{3}$

Add $8$ and $27$ .

$80 - \frac{35}{3}$

To write $80$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$80 \cdot \frac{3}{3} - \frac{35}{3}$

Combine $80$ and $\frac{3}{3}$ .

$\frac{80 \cdot 3}{3} - \frac{35}{3}$

Combine the numerators over the common denominator.

$\frac{80 \cdot 3 - 35}{3}$

Simplify the numerator.

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Multiply $80$ by $3$ .

$\frac{240 - 35}{3}$

Subtract $35$ from $240$ .

$\frac{205}{3}$

Find the Area Between the Curves y=16-x^2 , y=0 , x=-3 , x=2

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