mohon bantuannya ya???????

Maaf ya jawabannya ga ada di pg nya

Batas = 0 sampai 2
y = √x => x = y^2 ====> kurva kiri
y = 2 => x = 2^2 = 4 => kurva kanan

V = π integral (4^2 - (y^2)^2) dy
= π integral (16 - y^4) dy
= π [16y - 1/5 y^5] | batas 0 sampai 2
= π [(16 (2) - 1/5 (2)^5) - (0)]
= π [32 - 32/5]
= (160 - 32)/5 π
= 128/5 π
= 25 3/5 π