Penjelasan dengan langkah-langkah:

a) y = x² - 2 (parabola)

y = x (garis lurus)

diputar 360° di sumbu x

maka

y = y

x² - 2 = x

x² - x - 2 = 0

(x - 2)(x + 1) = 0

x1 = 2

x2 = -1

maka diperoleh daerah arsir ada pada selang

[ -1 2 ]

Maka rumus Volume benda putar terhadap sumbu x

= π ʃ (x²-2)² - (x)² dx [ -1 2 ]

= π ʃ x⁴ - 4x² + 4 - x² dx [ -1 2 ]

= π ʃ x⁴ - 5x² + 4 dx [ -1 2 ]

= π [ x⁵/5 - 5x³/3 + 4x ] [ -1 2 ]

= π [(2⁵/5 - 5(2)³/3 + 4(2))-((-1)⁵/5 - 5(-1)³/3 + 4(-1))]

= π (32/5 - 40/3 + 8) - ( -1/5 + 5/3 -4 )

= π (32/5 - 40/3 + 8) - (-38/15)

= π (32/5 - 40/3 + 8 + 38/15

= π (96/15 - 200/15 + 120/15 + 38/15)

= π (54/15)

= 18/5 π satuan

b) y = x + 2 ← p

2y = x + 2 = y = (x+2)/2 ← q

Batas [ -2 2 ]

diputar 360° di sumbu x

maka

p = q

x + 2 = (x+2)/2

2x + 4 = x + 2

= π ʃ p(x)² - q(x)² dx [ -2 2 ]

= π ʃ ( (x+2)²-(x+2/2)² dx [ -2 2 ]

= π ʃ (x²+4x+4)-((x²+4x+4)/4) dx [ -2 2 ]

= π x³/3 + 2x² + 4x - x³/12 - x²/2 - x [ -2 2 ]

= π x³/4 + 3x²/2 + 3x [ -2 2 ]

= π (2³/4 + 3(2)²/2 + 3(2)) - ((-2)³/4 + 3(-2)²/2 + 3(-2))

= π (2+6+6) - (-2 + 6 - 6 )

= π (14) - (-2)

= 16π satuan