1

Find the volume generated by revolving the area bounded by the curves $(x^2+4a^2)y=8a^3, 2y=x$ and $x=0$, about the y axis.

To find volume I am using this formula: $V=\pi\int [f(y)]^2dy$ where $f(y)= \sqrt{\frac{8a^3}{y}-4a^2}$

But how do I determine the limits of this integral?

user464068
  • 91

2 Answers2

1

Hint:

from the first equation we have $$ y=\frac{8a^3}{x^2+4a^2} $$

substituting $x=0$ we find $y=2a$

substituting $x=2y$ we find $y=a$

and you have to divide the interval for $y$ in two parts:

$0<y<a$ and $a<y<2a$ ( for $a>0$)

Emilio Novati
  • 62,675
1

If you make a drawing of the bounded area, you will see that, in the first quadrant, $y=x/2$ is an increasing line from $(0,0)$ and $y=\frac{8a^3}{x^2+4a^2}$ is a decreasing curve from $(0,2a)$. Their intersection point is $(2a,a)$. So, it is better to split the integral into two pieces in the following way $$V=\pi\int_{y=0}^a (2y)^2 dy+\pi\int_{y=a}^{2a} \left(\sqrt{\frac{8a^3}{y}-4a^2}\right)^2 dy.$$ Can you take it from here?

Robert Z
  • 145,942
  • Did you do this without tracing the curve or did you have an idea of what the curve would look like? – user464068 Jul 27 '17 at 10:05
  • @user464068 Yes, I made a sketch: $y=x/2$ is an increasing line from $(0,0)$ and $y=\frac{8a^3}{x^2+4a^2}$ is a decreasing curve from $(0,2a)$. Their intersection point is $(2a,a)$. – Robert Z Jul 27 '17 at 10:27