>>>Quiz<<<

Penjelasan dengan langkah-langkah:

1. y = 9 – x² dan y = x + 7

x + 7 = 9 – x²

x² + x – 2 = 0

(x + 2)(x – 1) = 0

x = -2 atau x = 1 ← batas integral

V = π ∫ ((9 – x²)² – (x + 7)²) dx

= π∫(x⁴ – 19x² – 14x + 32) dx

…………………………………..…1

= π[(1/5)x⁵ – (19/3)x³ – 7x² + 32x ]

………………………………….…-2

= π[(1/5) – (19/3) – 7 + 32 – ((1/5)(-2)⁵ – (19/3)(-2)³ – 7(-2)² + 32(-2))]

= π[(1/5) – (19/3) + 25 + (32/5) – (152/3) + 28 + 64]

= π[(33/5) – (171/3) + 117]

= (333/5)π

= 66 3/5 π satuan volume

2. y = √x dan y = 2

2 = √x

x = 2² = 4

integral dari x = 0 sampai x = 4

V = π ∫ (2² – √x²) dx = π ∫ (4 – x) dx

……….…….4

= π[4x – ½x²]

………….….0

= π[4.4 – ½.4² – 0]

= π(16 – 8)

= 8π satuan volume