Math / Calculus

QuestionEnter your answer as a fraction or as a number rounded to three decimal places. Find the area bounded by the graphs of $y=x^{2}$ and $y=32-x^{2}$ for $0 \leq x \leq 9$ . Area = $\square$

Studdy Solution

STEP 1

1. We are given two functions: $y = x^2$ and $y = 32 - x^2$ .
2. We need to find the area bounded by these two curves from $x = 0$ to $x = 9$ .
3. The area is calculated as the integral of the difference between the two functions over the given interval.

STEP 2

1. Determine the points of intersection of the two curves.
2. Set up the integral to find the area between the curves.
3. Evaluate the integral to find the area.

STEP 3

Find the points of intersection by setting the two equations equal to each other:
$x^2 = 32 - x^2$

STEP 4

Solve for $x$ :
$2x^2 = 32$ $x^2 = 16$ $x = \pm 4$
The points of intersection are $x = 4$ and $x = -4$ .

STEP 5

Set up the integral for the area between the curves from $x = 0$ to $x = 4$ :
$\text{Area} = \int_{0}^{4} [(32 - x^2) - x^2] \, dx$

STEP 6

Simplify the integrand:
$\text{Area} = \int_{0}^{4} (32 - 2x^2) \, dx$

STEP 7

Evaluate the integral:
$\text{Area} = \int_{0}^{4} 32 \, dx - \int_{0}^{4} 2x^2 \, dx$
$\text{Area} = \left[ 32x \right]_{0}^{4} - \left[ \frac{2}{3}x^3 \right]_{0}^{4}$

STEP 8

Calculate the definite integrals:
$\text{Area} = (32 \times 4 - 32 \times 0) - \left( \frac{2}{3}(4)^3 - \frac{2}{3}(0)^3 \right)$
$\text{Area} = 128 - \frac{2}{3}(64)$
$\text{Area} = 128 - \frac{128}{3}$
$\text{Area} = \frac{384}{3} - \frac{128}{3}$
$\text{Area} = \frac{256}{3}$
The area bounded by the graphs is:
$\boxed{\frac{256}{3}}$

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QuestionEnter your answer as a fraction or as a number rounded to three decimal places. Find the area bounded by the graphs of $y=x^{2}$ and $y=32-x^{2}$ for $0 \leq x \leq 9$ . Area = $\square$

Studdy Solution

STEP 1

1. We are given two functions: $y = x^2$ and $y = 32 - x^2$ .
2. We need to find the area bounded by these two curves from $x = 0$ to $x = 9$ .
3. The area is calculated as the integral of the difference between the two functions over the given interval.

STEP 2

1. Determine the points of intersection of the two curves.
2. Set up the integral to find the area between the curves.
3. Evaluate the integral to find the area.

STEP 3

Find the points of intersection by setting the two equations equal to each other:
$x^2 = 32 - x^2$

STEP 4

Solve for $x$ :
$2x^2 = 32$ $x^2 = 16$ $x = \pm 4$
The points of intersection are $x = 4$ and $x = -4$ .

STEP 5

Set up the integral for the area between the curves from $x = 0$ to $x = 4$ :
$\text{Area} = \int_{0}^{4} [(32 - x^2) - x^2] \, dx$

STEP 6

Simplify the integrand:
$\text{Area} = \int_{0}^{4} (32 - 2x^2) \, dx$

STEP 7

Evaluate the integral:
$\text{Area} = \int_{0}^{4} 32 \, dx - \int_{0}^{4} 2x^2 \, dx$
$\text{Area} = \left[ 32x \right]_{0}^{4} - \left[ \frac{2}{3}x^3 \right]_{0}^{4}$

STEP 8

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QuestionEnter your answer as a fraction or as a number rounded to three decimal places. Find the area bounded by the graphs of y=x2y=x^{2}y=x2 and y=32−x2y=32-x^{2}y=32−x2 for 0≤x≤90 \leq x \leq 90≤x≤9. Area = □\square□

Studdy Solution

STEP 1

1. We are given two functions: y=x2 y = x^2 y=x2 and y=32−x2 y = 32 - x^2 y=32−x2.2. We need to find the area bounded by these two curves from x=0 x = 0 x=0 to x=9 x = 9 x=9.3. The area is calculated as the integral of the difference between the two functions over the given interval.

STEP 2

1. Determine the points of intersection of the two curves.2. Set up the integral to find the area between the curves.3. Evaluate the integral to find the area.

STEP 3

Find the points of intersection by setting the two equations equal to each other:x2=32−x2 x^2 = 32 - x^2 x2=32−x2

STEP 4

Solve for x x x:2x2=32 2x^2 = 32 2x2=32 x2=16 x^2 = 16 x2=16 x=±4 x = \pm 4 x=±4The points of intersection are x=4 x = 4 x=4 and x=−4 x = -4 x=−4.

STEP 5

Set up the integral for the area between the curves from x=0 x = 0 x=0 to x=4 x = 4 x=4:Area=∫04[(32−x2)−x2] dx \text{Area} = \int_{0}^{4} [(32 - x^2) - x^2] \, dx Area=∫04​[(32−x2)−x2]dx

STEP 6

Simplify the integrand:Area=∫04(32−2x2) dx \text{Area} = \int_{0}^{4} (32 - 2x^2) \, dx Area=∫04​(32−2x2)dx

STEP 7

Evaluate the integral:Area=∫0432 dx−∫042x2 dx \text{Area} = \int_{0}^{4} 32 \, dx - \int_{0}^{4} 2x^2 \, dx Area=∫04​32dx−∫04​2x2dxArea=[32x]04−[23x3]04 \text{Area} = \left[ 32x \right]_{0}^{4} - \left[ \frac{2}{3}x^3 \right]_{0}^{4} Area=[32x]04​−[32​x3]04​

STEP 8

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QuestionEnter your answer as a fraction or as a number rounded to three decimal places. Find the area bounded by the graphs of $y=x^{2}$ and $y=32-x^{2}$ for $0 \leq x \leq 9$ . Area = $\square$

1. We are given two functions: $y = x^2$ and $y = 32 - x^2$ .
2. We need to find the area bounded by these two curves from $x = 0$ to $x = 9$ .
3. The area is calculated as the integral of the difference between the two functions over the given interval.

1. Determine the points of intersection of the two curves.
2. Set up the integral to find the area between the curves.
3. Evaluate the integral to find the area.

Find the points of intersection by setting the two equations equal to each other:
$x^2 = 32 - x^2$

Solve for $x$ :
$2x^2 = 32$ $x^2 = 16$ $x = \pm 4$
The points of intersection are $x = 4$ and $x = -4$ .

Set up the integral for the area between the curves from $x = 0$ to $x = 4$ :
$\text{Area} = \int_{0}^{4} [(32 - x^2) - x^2] \, dx$

Simplify the integrand:
$\text{Area} = \int_{0}^{4} (32 - 2x^2) \, dx$

Evaluate the integral:
$\text{Area} = \int_{0}^{4} 32 \, dx - \int_{0}^{4} 2x^2 \, dx$
$\text{Area} = \left[ 32x \right]_{0}^{4} - \left[ \frac{2}{3}x^3 \right]_{0}^{4}$