Question

Calc II Question

Find the volume of the solid obtained by rotating the region bonded bt the given curves about the specified line.

Y = e^-x
Y = 1
X = 2
About the Y = 2

Answer 1

`\displaystyle (\pi(5e^4+8e^2-1))/(2e^4)\approx9.526`

Explanation:

This can be solved with either the washer (easier) or the shell method (harder). For the disk/washer method, the slice is perpendicular to the axis of revolution, whereas, for the shell method, the slice is parallel to the axis of revolution. I'll show how to do it with both:

Shell Method (Horizontal Axis)

`\displaystyle V=2\pi\int^d_cr(y)h(y)\,dy`

Radius:
`r(y)=2-y` (distance from y=2 to x-axis)

Height:
`h(y)=2-(-\ln y)=2+\ln y` (
`y=e^(-x)` is the same as
`x=-\ln y`)

Bounds:
`[c,d]=[e^(-2),1]` (plugging x-bounds in gets you this)

Plugging in our integral, we get:

`\displaystyle V=2\pi\int^1_(e^(-2))(2-y)(2+\ln y)\,dy=(\pi(5e^4+8e^2-1))/(2e^4)\approx9.526`

Washer Method (Parallel to x-axis)

`\displaystyle V=\pi\int^b_a\biggr(R(x)^2-r(x)^2\biggr)\,dx`

Outer Radius:
`R(x)=2-e^(-x)` (distance between
`y=2` and
`y=e^(-x)`)

Inner Radius:
`r(x)=2-1=1` (distance between
`y=2` and
`y=1`)

Bounds:
`[a,b]=[0,2]`

Plugging in our integral, we get:

`\displaystyle V=\pi\int^2_0\biggr((2-e^(-x))^2-1^2\biggr)\,dx\\\\V=\pi\int^2_0\biggr((4-4e^(-x)+e^(-2x))-1\biggr)\,dx\\\\V=\pi\int^2_0(3-4e^(-x)+e^(-2x))\,dx\\\\V=\pi\biggr(3x+4e^(-x)-(1)/(2)e^(-2x)\biggr)\biggr|^2_0\\\\V=\pi\biggr[\biggr(3(2)+4e^(-2)-(1)/(2)e^(-2(2))\biggr)-\biggr(3(0)+4e^(-0)-(1)/(2)e^(-2(0))\biggr)\biggr]\\\\V=\pi\biggr[\biggr(6+4e^(-2)-(1)/(2)e^(-4)\biggr)-\biggr(4-(1)/(2)\biggr)\biggr]`

`\displaystyle V=\pi\biggr[\biggr(6+4e^(-2)-(1)/(2)e^(-4)\biggr)-(7)/(2)\biggr]\\\\V=\pi\biggr((5)/(2)+4e^(-2)-(1)/(2)e^(-4)\biggr)\\\\V=\pi\biggr((5)/(2)+(4)/(e^2)-(1)/(2e^4)\biggr)\\\\V=\pi\biggr((5e^4)/(2e^4)+(8e^2)/(2e^4)-(1)/(2e^4)\biggr)\\\\V=\pi\biggr((5e^4+8e^2-1)/(2e^4)\biggr)\\\\V=(\pi(5e^4+8e^2-1))/(2e^4)\approx9.526`

Use your best judgment when deciding on what method you use when visualizing the solid, but I hope this helped!