Graph question, is it correct please
2 answers:
You might be interested in
Answer:
V=25088π vu
Step-by-step explanation:
Because the curves are a function of "y" it is decided to take the axis of rotation as y
, according to the graph 1 the cutoff points of f(y)₁ and f(y)₂ are ±2
f(y)₁ = 7y²-28; f(y)₂=28-7y²
y=0; x=28-0 ⇒ x=28
x=0; 0 = 7y²-28 ⇒ 7y²=28 ⇒ y²= 28/7 =4 ⇒ y=√4 =±2
Knowing that the volume of a solid of revolution V=πR²h, where R²=(r₁-r₂) and h=dy then:
dV=π(7y²-28-(28-7y²))²dy ⇒dV=π(7y²-28-28+7y²)²dy = 4π(7y²-28)²dy
dV=4π(49y⁴-392y²+784)dy integrating on both sides
∫dV=4π∫(49y⁴-392y²+784)dy ⇒ solving ∫(49y⁴-392y²+784)dy
49∫y⁴dy-392∫y²dy+784∫dy =
V=4π( ) evaluated -2≤y≤2, or 2(0≤y≤2), also
⇒ V=25088π vu
