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Find the volume of the solid that results when the region enclosed by the given curve: $(11$$to14)$ $2ay2=x(x-a)2$ is revolved about $x$-axis ,$a>0$

1 Answer

  • 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 $\int \limits_a^b (-y) dx=\int \limits_a^b (-f(x))dx$
Step 1:
$2ay^2=x(x-a)^2$ is resolved about the x-axis.
$a > 0$ The curve is symmetric about x axes, passes through the orgin, does not exist for $x < 0$ and a loop is formed between $x=0\; and\; x=a$
Step 2:
Volume formed by rotating the loop
$V= \pi \int \limits_0^a y^2 dx$
$\quad=\large\frac{\pi}{2a} $$\int \limits_0^a x(x-a) ^2 dx$
$\quad= \large\frac{\pi}{2a} $$\int \limits_0^a (x^3-2ax^2 +a^2 x)dx$
$\quad= \large\frac{\pi}{2a} \bigg[\frac{x^4}{4}-\frac{2ax^3}{3}+\frac{a^2x^2}{2}\bigg]_0^a$
$\quad= \large\frac{\pi}{2a} \bigg[\frac{a^4}{4}-\frac{2ax^4}{3}+\frac{a^4}{2}\bigg]$
$\quad= \large\frac{\pi}{24a}$$ \bigg[3a^4-8a^4+6a^4\bigg]$
$\quad=\large\frac{\pi a^3}{24} $$units$
answered Aug 16, 2013 by meena.p

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