How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by #y=f(x)=3x-x^2# and x axis revolved about the x=-1?

Question

Isaiah Allen · Accepted Answer

#V=(45pi)/2 units^3#

Imagine a cylindrical shell as a rectangular prism with width #f(x)#, length #2pir#, and thickness #deltax#. The reason for the length being #2pir# is that if we unravel a cylindrical shell into a rectangular prism, the length corresponds to the circumference of the circular cross-section of the original cylinder.

The volume of the cylindrical shell is width x length x thickness (height)

Now width #=f(x)=3x-x^2#

#:.# Volume of shell#=2pir(3x-x^2)deltax#

If we sketch the parabola #y=3x-x^2# we can see that the region bound by this parabola and the x-axis resides in the domain #0<=x<=3#

Also, the radius of the cylindrical shell is considered to be the distance from the axis of revolution #x=-1# and the edge of the shell (i.e. at a position #x# within the domain #0<=x<=3#)

Hence, r#=1+x#

#:.#Volume of shell#=2pi(1+x)(3x-x^2)deltax#

i.e. the change in volume, #deltav=2pi(1+x)(3x-x^2)deltax#

To find the volume, we limit the thickness of the shells, and find their summation within the domain #0<=x<=3#

#V=lim_(deltax->0)sum_(x=0)^3(2pi(1+x)(3x-x^2)deltax)#

#=2piint_0^3(1+x)(3x-x^2)dx#

#=2piint_0^3(3x+2x^2-x^3)dx#

#=2pi[3/2x^2+2/3x^3-1/4x^4]_0^3#

#=2pi[45/4]#

#=(45pi)/2 units^3#

Isaiah Allen · Answer

undefined To use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by $y = f(x) = 3x - x^2$ and the x-axis, revolved about the line $x = -1$, follow these steps:

1. Identify the region of interest, which in this case is the area under the curve $y = f(x) = 3x - x^2$ bounded by the x-axis and the y-axis.

2. Determine the limits of integration. Since we're revolving about the line $x = -1$, the limits of integration will be from the smallest x-value of the region to the largest x-value of the region. To find these limits, set $3x - x^2 = 0$ and solve for x.

$3x - x^2 = 0$

$x(3 - x) = 0$

$x = 0$ and $x = 3$

So, the limits of integration are from $x = 0$ to $x = 3$.

3. Set up the integral for the volume using the formula for cylindrical shells:

$V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx$

where $a$ and $b$ are the lower and upper limits of integration respectively.

4. Substitute the function $f(x) = 3x - x^2$ into the integral.

$V = 2\pi \int_{0}^{3} x \cdot (3x - x^2) \, dx$

5. Simplify the expression and integrate.

6. Evaluate the integral to find the volume of the solid obtained by rotating the region bounded by $y = f(x) = 3x - x^2$ and the x-axis, revolved about the line $x = -1$.