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Find the volume of the solid that results when the region enclosed by the given curve: $\;y=x^{3},x=0,y=1$ is revolved about the $y$-axis.

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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:
Volume of solid generated by resolving the area between $y=x^3 ,y=0,y=1$ about y-axis
$V= \pi \int \limits_0^1 x^2 dy$
$\quad= \pi \int \limits_0^1 y^{\large\frac{2}{3}} dy$
$\quad=\large\frac{3}{5} $$ \pi \bigg[y^{\large\frac{5}{3}} \bigg]_0^1$
$\quad=\large\frac{3 \pi}{5}$


answered Aug 16, 2013 by meena.p
edited Aug 16, 2013 by meena.p

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