Calculus Examples

$y = 2 x^{2}$ , $y = 0$ , $x = 2$

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 $A = π r^{2}$ .

$V = π \int_{0}^{2} {(f (x))}^{2} d x$ where $f (x) = 2 x^{2}$

Step 2

Simplify the integrand.

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Step 2.1

Apply the product rule to $2 x^{2}$ .

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

Step 2.2

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

$V = 4 {(x^{2})}^{2}$

Step 2.3

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

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Step 2.3.1

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

$V = 4 x^{2 \cdot 2}$

Step 2.3.2

Multiply $2$ by $2$ .

$V = 4 x^{4}$

Step 3

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

$V = π (4 \int_{0}^{2} x^{4} d x)$

Step 4

Move $4$ to the left of $π$ .

$V = 4 π \int_{0}^{2} x^{4} d x$

Step 5

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

$V = 4 π \frac{1}{5} x^{5}]_{0}^{2}$

Step 6

Simplify the answer.

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Step 6.1

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

$V = 4 π \frac{x^{5}}{5}]_{0}^{2}$

Step 6.2

Substitute and simplify.

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Step 6.2.1

Evaluate $\frac{x^{5}}{5}$ at $2$ and at $0$ .

$V = 4 π ((\frac{2^{5}}{5}) - \frac{0^{5}}{5})$

Step 6.2.2

Simplify.

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Step 6.2.2.1

Raise $2$ to the power of $5$ .

$V = 4 π (\frac{32}{5} - \frac{0^{5}}{5})$

Step 6.2.2.2

Raising $0$ to any positive power yields $0$ .

$V = 4 π (\frac{32}{5} - \frac{0}{5})$

Step 6.2.2.3

Cancel the common factor of $0$ and $5$ .

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Step 6.2.2.3.1

Factor $5$ out of $0$ .

$V = 4 π (\frac{32}{5} - \frac{5 (0)}{5})$

Step 6.2.2.3.2

Cancel the common factors.

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Step 6.2.2.3.2.1

Factor $5$ out of $5$ .

$V = 4 π (\frac{32}{5} - \frac{5 \cdot 0}{5 \cdot 1})$

Step 6.2.2.3.2.2

Cancel the common factor.

$V = 4 π (\frac{32}{5} - \frac{5 \cdot 0}{5 \cdot 1})$

Step 6.2.2.3.2.3

Rewrite the expression.

$V = 4 π (\frac{32}{5} - \frac{0}{1})$

Step 6.2.2.3.2.4

Divide $0$ by $1$ .

$V = 4 π (\frac{32}{5} - 0)$

Step 6.2.2.4

Multiply $- 1$ by $0$ .

$V = 4 π (\frac{32}{5} + 0)$

Step 6.2.2.5

Add $\frac{32}{5}$ and $0$ .

$V = 4 π (\frac{32}{5})$

Step 6.2.2.6

Combine $\frac{32}{5}$ and $4$ .

$V = \frac{32 \cdot 4}{5} \cdot π$

Step 6.2.2.7

Multiply $32$ by $4$ .

$V = \frac{128}{5} \cdot π$

Step 6.2.2.8

Combine $\frac{128}{5}$ and $π$ .

$V = \frac{128 π}{5}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$V = \frac{128 π}{5}$

Decimal Form:

$V = 80.42477193 \dots$

Step 8