This question is incomplete, the complete question is;
Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y2 = 2x, x = 2y; about the y-axis
Answer:
V = π (512/15)
Explanation:
Given that;
region of rotation
y² = 2x, x = 2y
Region is rotated about y-axis as shown in the image
for the point of intersection,
y²/2 = 2y
y² - 4y = 0
y(y-4) = 0
∴ y = 0, y = 4
so the region lies in 0 ≤ y ≤ 4
Now cross section area of washer is
A(y) = π(outer radius)² = π(inner radius)²
A(y) = π(2y)² - π(y²/2)²
A(y) = π(4y²) - π(y⁴/4)
A(y) = π(4y² - (y⁴/4))
now volume of the solid of revolution is
V = ⁴∫₀ A(y) dy
V = ⁴∫₀ π(4y² - (y⁴/4))dy
V = π {4⁴∫₀ y² - 1/4⁴∫₀y⁴ dy }
V = π { 4/3 [y³]₀⁴ - 1/20 [y⁵]₀⁴ }
V = π { 4/3 [4]₀⁴ - 1/20 [4]₀⁴ }
V = π { 4/3 [64]₀⁴ - 1/20 [1024]₀⁴ }
V = π { 256/3 - 1024/20 }
V = π { (5120 - 3072) / 60 }
V = π (512/15)