Find the volume of the solid obtained by revolving the area enclosed by the curve $27ay^{2} = 4(x-2a)^{3} , x$ axis and parabola $y^{2} = 4ax$ about the $x$ axis.

I am not able to find the area enclosed by both curves. I know that $27ay^{2} = 4(x-2a)^{3}$ is symmetric about $x$ axis, I know how to draw both curves but don't know if parabola should be drawn above or below the other curve. I tried to find the maximum value of $y$ for loop by differentiating it with respect to $x$ but getting only its minimum value. Also I couldn't find their intersection point.

So how can I have the idea about the area enclosed between them$?$


You can plug $y^2 = 4ax$ into $27ay^2 = 4(x-2a)^3$. You will end up with: $$27a\cdot4ax=4(x-2a)^3$$ Solving this for x yields: $$x=8a$$

This is your intersection point btw.

enter image description here

This has been drawn using $a = 3$. By looking at your equation you can see that the first equation is $0$ for $x = 2a$.

Using the formula for volume:

$$\int_{x_1}^{x_2}\pi f(x)^2 dx$$ We need to split this into two parts from $0$ to $2a$ and from $2a$ to $8a$:

$$\int_{x_1}^{x_2}\pi f(x)^2 dx = \int_{0}^{2a}\pi f_1(x)^2 dx + \int_{2a}^{8a}\pi (f_1(x)-f_2(x))^2 dx$$

I am sure you can solve this yourself.


First you shall determine the domain of definition of each curve.

For the first we have $$ 0 \le y^{\,2} = {{4\left( {x - 2a} \right)^{\,3} } \over {27a}}\quad \Rightarrow \quad \left\{ {\matrix{ {x < 2a} & {a < 0} \cr {2a < x} & {0 < a} \cr } } \right. $$ and for the second $$ x = {{y^{\,2} } \over {4a}}\quad \Rightarrow \quad \left\{ {\matrix{ {x < 0} & {a < 0} \cr {0 < x} & {0 < a} \cr } } \right. $$ So if $a$ changes sign, we just have a reflection around the $y$ axis, We can consider only the case $0 < a$.

Then let's determine when the first curve is over the second ($y_2 \le y_1$).
Since we have $$ \left\{ \matrix{ 27ay^{\,2} = 4\left( {x - 2a} \right)^{\,3} \hfill \cr y^{\,2} = 4ax \hfill \cr} \right.\quad \Leftrightarrow \quad \left\{ \matrix{ 27ay^{\,2} = 4\left( {x - 2a} \right)^{\,3} \hfill \cr 27ay^{\,2} = 108a^{\,2} x \hfill \cr} \right. $$ that means $$ \eqalign{ & 0 \le \left( {x - 2a} \right)^{\,3} - 27a^{\,2} x = \cr & = \left( {x - 2a} \right)^{\,3} - 27a^{\,2} \left( {x - 2a} \right) - 54a^{\,3} = \cr & = \left( {x/a - 2} \right)^{\,3} - 27\left( {x/a - 2} \right) - 54 = \cr & = \left( {x/a - 8} \right)\left( {x/a + 1} \right)^{\,2} \cr} $$

So, actually we have only the $y_2 = 2 \sqrt{ax}$ in $0 < x < 2a$, and $y_1 < y_2$ for $ 2a < x < 8a$.

After that, I think you can easily split the integral, and proceed by yourself.


For this particular revolved shape, it is cleaner to compute the volume with the 'cylinder integration', since it requires only one set of boundaries, i.e.



$$x_1(y)=\left(\frac{27a}{4}\right)^{1/3}y^{2/3} + 2a $$ $$x_2(y)=\frac{1}{4a}y^2 $$

The upper integral limit is $b=\sqrt{32}a$, which is the y-coordinate at which the two curves intersect. Then,

$$V = 2\pi \int_0^{\sqrt{32}a} \left[ \left(\frac{27a}{4}\right)^{1/3} y^{5/3} + 2ay - \frac{1}{4a}y^3 \right] dy $$

The three polynomial integrands can in turn be integrated piecewise and they sum up to (144+64-128)$\pi a^3$ = 80$\pi a^3$.

  • $\begingroup$ Answer is given 80$\pi a^{3}$ $\endgroup$ – Mathsaddict Aug 5 '19 at 16:11
  • $\begingroup$ Just realized the solid revolves around x, not y. So, the corresponding integral $V/\pi=4a\int_0^{8a} xdx - 4/(27a)\int_{2a}^{8a}(x-2a)^3 dx = 128a^3-48a^3=80a^3$. $\endgroup$ – Quanto Aug 5 '19 at 17:32

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