Answer:
The volume of the resulting solid is π/2 cubic units.
Step-by-step explanation:
Please refer to the diagram below.
The shell method is given by:
[tex]\displaystyle V = 2\pi \int _a ^b r(x) h(x)\, dx[/tex]
Where the representative rectangle is parallel to the axis of revolution, r(x) is the distance from the axis of revolution to the center of the rectangle, and h(x) is the height of the rectangle.
From the diagram, we can see that r(x) = (2 - x) and that h(x) is simply y. The limits of integration are from a = 0 to b = 1. Therefore:
[tex]\displaystyle V = 2\pi \int_0^1\underbrace{\left(2-x\right)}_{r(x)}\underbrace{\left(x - x^2\right)}_{h(x)}\, dx[/tex]
Evaluate:
[tex]\displaystyle \begin{aligned} V&= 2\pi \int_0 ^1 \left(2x-2x^2-x^2+x^3\right) \, dx\\ \\ &= 2\pi\int _0^1 x^3 -3x^2 + 2x \, dx \\ \\ &= 2\pi\left(\frac{x^4}{4} - x^3 + x^2 \Bigg|_0^1\right) \\ \\ &= 2\pi \left(\frac{1}{4} - 1 + 1 \right) \\ \\ &= \frac{\pi}{2}\end{aligned}[/tex]
The volume of the resulting solid is π/2 cubic units.
Answer:
pi/2
Step-by-step explanation:
I always like to draw an illustration for these problems.
For shells method think volume of cylinder=2pi×r×h
Integrate(2pi(2-x)(x-x^2) ,x=0...1)
Multiply
Integrate(2pi(2x-2x^2-x^2+x^3 ,x=0...1)
Combine like terms
Integrate(2pi(2x-3x^2+x^3) ,x=0...1)
Begin to evaluate
2pi(2x^2/2-3x^3/3+x^4/4) ,x=0...1
2pi(x^2-x^3+x^4/4), x=0...1
2pi(1-1+1/4)
2pi/4
pi/2
