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Find the volume of the solid that results when the region enclosed by the given curve: $(11$$to14)$$\large\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is revolved about major axis $a>b>0.$

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Toolbox:
  • Area bounded by the curve $t=f(x),$ the x-axis and the ordinates $x=a,x=b$ is $\int \limits_a^b f(x) dx $ or $ \int \limits _a^b y dx $
  • If the curve lies below the x-axis for $a \leq x \leq b,$ then the area is $\int \limits_a^b (-y) dx=\int \limits_a^b (-f(x))dx$
Volume of solid generated by rotating $\large\frac{x^2}{a^2}+\large\frac{y^2}{b^2}$$=1$ about the major axis (x-axis since a>b) is
$V=\pi\int\limits_{-a}^a y^2 dx=2\pi\int\limits _0^a y^2 dx$
$\qquad=2\pi\int\limits_0^a b^2(1-\large\frac{x^2}{a^2})$$dx$
$\qquad=2\pi[b^2x-\large\frac{b^2}{a^2}.\frac{x^3}{3}\bigg]_0^a$
$\qquad=2\pi\bigg[ab^2-\large\frac{b^2a^3}{a^23}\bigg]$
$\qquad=\large\frac{4\pi}{3}$$ab^2$$units$
answered Aug 16, 2013 by meena.p
 

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