Question

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle.
x = 28 − 7y2,
x = 7y2 − 28

Answer 1

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( $49\frac{y^{5} }{5}-392\frac{y^{3}}{3}+784y$ ) evaluated -2≤y≤2, or 2(0≤y≤2), also

$V=8\pi(49\frac{2^{5} }{5}-392\frac{2^{3} }{3}+784.2)$  ⇒ V=25088π vu