Calculus Examples

$y = x^{2}$ , $y = 1$

Step 1

To find the volume of the solid, first define the area of each slice then integrate across the range. The area of each slice is the area of a circle with radius $f (x)$ and $A = π r^{2}$ .

$V = π \int_{- 1}^{1} {(f (x))}^{2} - {(g (x))}^{2} d x$ where $f (x) = 1$ and $g (x) = x^{2}$

Step 2

Simplify each term.

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Step 2.1

One to any power is one.

$V = 1 - {(x^{2})}^{2}$

Step 2.2

Multiply the exponents in ${(x^{2})}^{2}$ .

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Step 2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$V = 1 - x^{2 \cdot 2}$

Step 2.2.2

Multiply $2$ by $2$ .

$V = 1 - x^{4}$

Step 3

Split the single integral into multiple integrals.

$V = π (\int_{- 1}^{1} d x + \int_{- 1}^{1} - x^{4} d x)$

Step 4

Apply the constant rule.

$V = π (x]_{- 1}^{1} + \int_{- 1}^{1} - x^{4} d x)$

Step 5

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

$V = π (x]_{- 1}^{1} - \int_{- 1}^{1} x^{4} d x)$

Step 6

By the Power Rule, the integral of $x^{4}$ with respect to $x$ is $\frac{1}{5} x^{5}$ .

$V = π (x]_{- 1}^{1} - (\frac{1}{5} x^{5}]_{- 1}^{1}))$

Step 7

Simplify the answer.

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Step 7.1

Combine $\frac{1}{5}$ and $x^{5}$ .

$V = π (x]_{- 1}^{1} - (\frac{x^{5}}{5}]_{- 1}^{1}))$

Step 7.2

Substitute and simplify.

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Step 7.2.1

Evaluate $x$ at $1$ and at $- 1$ .

$V = π ((1) + 1 - (\frac{x^{5}}{5}]_{- 1}^{1}))$

Step 7.2.2

Evaluate $\frac{x^{5}}{5}$ at $1$ and at $- 1$ .

$V = π (1 + 1 - (\frac{1^{5}}{5} - \frac{{(- 1)}^{5}}{5}))$

Step 7.2.3

Simplify.

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Step 7.2.3.1

Add $1$ and $1$ .

$V = π (2 - (\frac{1^{5}}{5} - \frac{{(- 1)}^{5}}{5}))$

Step 7.2.3.2

One to any power is one.

$V = π (2 - (\frac{1}{5} - \frac{{(- 1)}^{5}}{5}))$

Step 7.2.3.3

Raise $- 1$ to the power of $5$ .

$V = π (2 - (\frac{1}{5} - \frac{- 1}{5}))$

Step 7.2.3.4

Move the negative in front of the fraction.

$V = π (2 - (\frac{1}{5} + \frac{1}{5}))$

Step 7.2.3.5

Multiply $- 1$ by $- 1$ .

$V = π (2 - (\frac{1}{5} + 1 (\frac{1}{5})))$

Step 7.2.3.6

Multiply $\frac{1}{5}$ by $1$ .

$V = π (2 - (\frac{1}{5} + \frac{1}{5}))$

Step 7.2.3.7

Combine the numerators over the common denominator.

$V = π (2 - \frac{1 + 1}{5})$

Step 7.2.3.8

Add $1$ and $1$ .

$V = π (2 - \frac{2}{5})$

Step 7.2.3.9

To write $2$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$V = π (2 \cdot \frac{5}{5} - \frac{2}{5})$

Step 7.2.3.10

Combine $2$ and $\frac{5}{5}$ .

$V = π (\frac{2 \cdot 5}{5} - \frac{2}{5})$

Step 7.2.3.11

Combine the numerators over the common denominator.

$V = π (\frac{2 \cdot 5 - 2}{5})$

Step 7.2.3.12

Simplify the numerator.

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Step 7.2.3.12.1

Multiply $2$ by $5$ .

$V = π (\frac{10 - 2}{5})$

Step 7.2.3.12.2

Subtract $2$ from $10$ .

$V = π (\frac{8}{5})$

Step 7.2.3.13

Combine $π$ and $\frac{8}{5}$ .

$V = \frac{π \cdot 8}{5}$

Step 7.2.3.14

Move $8$ to the left of $π$ .

$V = \frac{8 π}{5}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$V = \frac{8 π}{5}$

Decimal Form:

$V = 5.02654824 \dots$

Step 9