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y= -2x+4 ⇒y2
⇒mencari titik potong
y1 = -x²+4
0 = -x²+4 x= 0
x²-4 = 0 x= -0²+4
(x+2)(x-2) = 0 x= 0+4
x= -2 atau x= 2 x= 4
y2 = -2x+4
0 = -2x+4 x= 0
2x = 4 x= -2(0)+4
x = 2 x= 4
⇒karna diputar mengelilingi sumbu y,maka batasnya 0 dan 4
maka rubah y=... menjadi x=...
y= -x²+4 y= -2x+4
x²= 4-y 2x= 4-y
x= 2-1/2y
x²= 4-1/4y²
f(y)²= y1-y2
= 4-y-(4-1/4y²)
= 4-y-4+1/4y²
= y-1/4y²
⇒cari volume
V= π ₀⁴∫ (y-1/4y²) dy
V= π [1/2y²-1/12y³]₀⁴
V= π [(1/2(4)²-1/12(4)³) -(0)]
V= π [8-64/12]
V= π [32/12]
V= 8/3π satuan volume⇒ 2 2/3(C)