Question

The volume generatred by revolving the area bounded by the parabola $y^2=8x$ and the line $x=2$ about $y=$axis is

• A. $\frac{128\pi }{5}$
• B. $\frac{5}{128\pi }$
• C. $\frac{127}{5\pi }$
• D. None of above

Answer 1

The correct option is A $\frac{128\pi }{5}$


Volume $={\int }_0^2\left(2\pi xdx\right)\times 2y$
$={\int }_0^{2}4\pi x\sqrt{8x}dx$
$=8\sqrt{2}\times \pi \times \frac{2}{5}[2{]}^{5/2}$
Volume $=\frac{128\pi }{5}$

Method 2:


Total volume of the cylindrical shaped solid formed by the revolution of the surface $POQ$ about $y-$axis
$=\pi r^{2}h=\pi \left(2{)}^2\right(8)=32\pi$
Now, volume of the solid formed by the revolution of the curve $POQ$ about $y-$axis
$=2$ volume of solid formed by curve $PO$ about $y-$axis
$=2.\int \pi x^{2}dy=2.{\int }_{y=0}^4\pi \left(\frac{y^4}{64}\right)dy$
$=\frac{32\pi }{5}$
So, Required volume $=32\pi -\frac{32\pi }{5}=\frac{128\pi }{5}$