Pre-Algebra Examples

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\begin{matrix} h = & 2 r = & 6 \end{matrix}

$\begin{array}{r} h = & 2 \\ r = & 6 \end{array}$

Step 1

The surface area of a cylinder is equal to the sum of the areas of the bases each having an area of

π \cdot r^{2}

$π \cdot r^{2}$ , plus the area of the side. The area of the side is equal to the area of a rectangle with length

2 π r

$2 π r$ (the circumference of the base) times the height.

2 π \cdot {(r a d i u s)}^{2} + 2 π \cdot (r a d i u s) \cdot (h e i g h t)

$2 π \cdot {(r a d i u s)}^{2} + 2 π \cdot (r a d i u s) \cdot (h e i g h t)$

Step 2

Substitute the values of the radius

r = 6

$r = 6$ and height

h = 2

$h = 2$ into the formula. Pi

π

$π$ is approximately equal to

3.14

$3.14$ .

2 (π) {(6)}^{2} + 2 (π) (6) (2)

$2 (π) {(6)}^{2} + 2 (π) (6) (2)$

Step 3

Simplify each term.

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Step 3.1

Raise

6

$6$ to the power of

2

$2$ .

2 π \cdot 36 + 2 (π) (6) (2)

$2 π \cdot 36 + 2 (π) (6) (2)$

Step 3.2

Multiply

36

$36$ by

2

$2$ .

72 π + 2 (π) (6) (2)

$72 π + 2 (π) (6) (2)$

Step 3.3

Multiply

6

$6$ by

2

$2$ .

72 π + 12 π \cdot 2

$72 π + 12 π \cdot 2$

Step 3.4

Multiply

2

$2$ by

12

$12$ .

72 π + 24 π

$72 π + 24 π$

72 π + 24 π

$72 π + 24 π$

Step 4

Add

72 π

$72 π$ and

24 π

$24 π$ .

96 π

$96 π$

Step 5

The result can be shown in multiple forms.

Exact Form:

96 π

$96 π$

Decimal Form:

301.59289474 \dots

$301.59289474 \dots$

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