Solids of Revolution

6.3

• using the same idea of Volumes of integrals

  • But now revolving it around an axis instead of perpendicular

• Solids of revolution

  • Discs

    • Ex

      • A. $y=\sin x,x=\frac{\pi }{2}and,y=0$

        • looking at the cross sections we know that the cross section is a disc and the radius is sin(x) $\therefore$ $A(x)=\pi {\sin }^{2}x$
          • $V={\int }_0^{\frac{\pi }{2}}A(x)dx⟹\pi {\int }_0^{\pi /2}{\sin }^{2}xdx$
            • we can’t integrate that :( so we use a trig identity
              • $\cos 2x=2{\cos }^{2}x-1⟹{\cos }^{2}x=\frac{{\cos }_{2}x+1}{2}⟹{\sin }^{2}x=1-{\cos }^{2}x=\frac{1-\cos 2x}{2}$
                • $\pi {\int }_0^{\frac{\pi }{2}}\frac{1}{2}dx$

      • B. $y=\sqrt{x},y=1,x=0$

        • to find volume about the y axis
          • Again we get a disc with the radius $y^2$
            • $A(y)=(y^2{)}^2\ast \pi =\pi y^4$
              • => $V={\int }_0^1\pi y^{4}dy⟹\frac{\pi }{5}$

  • Washers

    • EX

      • A, volume of the solid between y=x and y=x^3 along the lines of y=3 x=2

        • to find the integral we must find the area, and that area is defined by the “gap” minus the area
          • $A(x)=\pi (3-x^3{)}^2-\pi (3-x{)}^2$
            • $V={\int }_0^1\pi (x^6-6x^3-x^2+6x)dx$
        • now about the y axis
          • $A(y)=\pi (2-y{)}^2-(2-y^1/)$

      • 3d washers (torus)

        • to setup the integral for the volume of a torus we will need to know small radius of the circle inside then the R radius for the volume of revolution it self
          • we can solve that using the circle equation $x^2+y^2=r^2$
            • thus we can deduce that the equation is $(x-R{)}^2+y^2=r^2$ big R is the vertical translation to the right
              • => $x=R\pm \sqrt{r^2-y^2}$
                • To get the translation of the integral we need to know what our outer radius and inner radius is, we know that R is from the center point to the middle of the circle, so we can use minus R to get the inner and plus R to get the outer
                  • $V={\int }_{-r}^r\pi (R+\sqrt{r^2-y^2}{)}^2-\pi (R-\sqrt{r^2-y^2}{)}^{2}dx$

  • Shell method

    • Def.

      • We use cylinders that its area of 2$\pi$x in length and f(x) in height

        • now we can use an integral
          • $V=2\pi {\int }_a^{b}xf(x)dx⟹{\int }_a^{b}A(x)dx$

    • Ex.

      • 1. find the volume of the solid obtained by rotating about the y-axis the region bounded by curves

        • $y=3x^2-x^3$ and y =0
          • $V={\int }_0^{3}2\pi x(3x^2-x^3)dx$
            • => $2\pi {\int }_0^{3}3x^3-x^{4}dx$
              • => $2\pi \left(\frac{3}{4}{x}^4-\frac{x^5}{5}\right)$

      • 2. same thing but with two functions

        • find the difference between the two then use the integral wawawawaaw