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Determining the Volume of a Solid of Revolution

Key Questions

  • 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#

Questions