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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$?$

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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.

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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.

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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.

$$V=2\pi\int_0^{b}y[x_2(y)-x_1(y)]dy$$

where

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

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$\pi a^3$.

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  • $\begingroup$ 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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