Determining the Volume of a Solid of Revolution

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Volumes of Revolution - Disk/Washers Example 1
4:35 — by patrickJMT

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Key Questions

  • The volume of a solid of revolution for a curve #y = f(x)# rotated around the #x#-axis can be found through: #V = pi int_a^b y^2 dx#
    To solve this integral you must find the bounds #a# and #b#.

    For example, the curve #y=x^2 + 2# is rotated between the bounds #x = 1# and #x = 2# is rotated about the #x#-axis. Find the volume of the solid formed.
    Since #y = x^2 + 2#
    #y^2 = x^4 + 4x^2 + 4#
    #:. V= pi int_1^2 x^4 + 4x^2 + 4 dx#
    #= pi [ (x^5)/(5) + (4x^3)/(3) +4x]_1^2#
    #= (431pi)/5# or #270.81#(2dp) units cubed

    When the curve is rotated around the #y#-axis the volume of the solid is given by: #V = pi int_e^d x^2 dx#
    The function must be in the form where #x# is a function of #y# and the bounds #d# and #e# must be read off the #y#-axis.

  • The short and sweet version of the answer is:
    #(pi)*int_0^1((4-x^2)^2-(4-x)^2)dx#=

    #1.2pi# or approximately #3.769911 . . .#

    This is the region bounded by the points #(0,0)# and #(1,1)# since those points lie on both the graph of the line (#g(x)=x#) and the parabola (#(f(x)=x^2#). We will take this region and subdivide it into vertical regions that are one can picture as tiny rectangle. When the rectangles are spun about the line #y=4#, they will form tiny washers. If there are #n# subintervals, then the width of each subinterval is #Delta x=1/n#. The outer radius (distance from the function farther away from the line of rotation to the line of rotation) is #(4-f(x))#. The inner radius (distance from the function nearer to the line of rotation to the line of rotation) is #(4-g(x))

    The usual setup for washers is:

    #*lim_(nrarroo)sum_(i=1)^n((piR(x))^2-(pir(x))^2)*(Deltax)# which becomes

    #(pi)*int_a^b((piR(x)^2-(pir(x))^2)*(dx)# which for our functions equals:

    #int_0^1(pi(4-f(x))^2-pi(4-g(x))^2)dx#=#(pi)*int_0^1((4-x^2)^2-(4-x)^2)dx#=

    #1.2pi# or approximately #3.769911 . . .#

  • A cone with base radius #r# and height #h# can be obtained by rotating the region under the line #y=r/hx# about the x-axis from #x=0# to #x=h#.
    By Disk Method,
    #V=pi int_0^h(r/hx)^2 dx={pi r^2}/{h^2}int_0^hx^2 dx#
    by Power Rule,
    #={pir^2}/h^2[x^3/3]_0^h={pir^2}/{h^2}cdot h^3/3=1/3pir^2h#

  • If the radius of its circular cross section is #r#, and the radius of the circle traced by the center of the cross sections is #R#, then the volume of the torus is #V=2pi^2r^2R#.

    Let's say the torus is obtained by rotating the circular region #x^2+(y-R)^2=r^2# about the #x#-axis. Notice that this circular region is the region between the curves: #y=sqrt{r^2-x^2}+R# and #y=-sqrt{r^2-x^2}+R#.

    By Washer Method, the volume of the solid of revolution can be expressed as:
    #V=pi int_{-r}^r[(sqrt{r^2-x^2}+R)^2-(-sqrt{r^2-x^2}+R)^2]dx#,
    which simplifies to:
    #V=4piR\int_{-r}^r sqrt{r^2-x^2}dx#
    Since the integral above is equivalent to the area of a semicircle with radius r, we have
    #V=4piRcdot1/2pi r^2=2pi^2r^2R#

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Applications of Definite Integrals