Given that the function graphed is ƒ(x), what is ƒ(1)?
1 answer:
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Answer:
V = 34,13*π cubic units
Step-by-step explanation: See Annex
We find the common points of the two curves, solving the system of equations:
y² = 2*x x = 2*y ⇒ y = x/2
(x/2)² = 2*x
x²/4 = 2*x
x = 2*4 x = 8 and y = 8/2 y = 4
Then point P ( 8 ; 4 )
The other point Q is Q ( 0; 0)
From these two points, we get the integration limits for dy ( 0 , 4 )are the integration limits.
Now with the help of geogebra we have: In the annex segment ABCD is dy then
V = π *∫₀⁴ (R² - r² ) *dy = π *∫₀⁴ (2*y)² - (y²/2)² dy = π * ∫₀⁴ [(4y²) - y⁴/4 ] dy
V = π * [(4/3)y³ - (1/20)y⁵] |₀⁴
V = π * [ (4/3)*4³ - 0 - 1/20)*1024 + 0 )
V = π * [256/3 - 51,20]
V = 34,13*π cubic units
