tolong bantu jawab soal no 5 dan 6 tentang volume benda putar dengan cara penyelesaiannya

jawab
(5)

y = x²
y = 2 - x
x²= 2 - x
x² + x - 2 = 0
(x + 2)(x - 1)=0
x  = - 2  atau x= 1

y₁ = x² → y² = x⁴
y₂ = 2-x   → y² = (2- x)² = 4 - 4x + x²
Grafik  garis diatas kurva

Volume  = π ₋₂¹ ∫(y₂² - y₁²) dx
= π   ₋₂¹∫ (x² -4x + 4 - x⁴) dx
= π [ 1/3 x³ - 2x² + 4x - 1/5  x⁵]¹₋₂
= π [ 1/3(1+8) - 2(1 - 4) + 4(1+2) - 1/5 (1 +32)]
= π ( 3 + 6 + 12 - 33/5)
= 14 ²/₅ π

(6)
y₁ = x²
y₂ = -x² + 2
x²  = -x² + 2
2x² - 2=0
2(x² -1)=0
x = 1  atau x = -1

y₁² = x⁴
y₂² = (-x² +2)² = x⁴ - 4x² + 4
grafik  y₂ diatas y₁
batas integral x = -1 sd x = 1

V=  π ₋₁¹ ∫ (y₂² - y₁²) dx
V = π ₋₁¹∫ (x⁴ - 4x² + 4 - x⁴) dx
V = π ₋₁¹ ∫ ( -4x² + 4 ) dx
V = π [ -4/3 x³ + 4x]¹₋₁
V = π [ -4/3 (1 +1) + 4(1 +1)]
V = π [ - 8/3 + 8 ]
V = (16/3) π