Welcome to our community

Be a part of something great, join today!

The volume of the solid generated by the revolution

ksananthu

New member
Jul 14, 2013
5
find the volume of the solid generated by the revolution of the curve
$y^2 (2 a - x) = x^3$ about its asymptote.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: the volume of the solid generated by the revolution

Have you identified the axis of revolution, i.e., the asymptote, and have you observed which method of computing the solid is the most practical and why?
 

ksananthu

New member
Jul 14, 2013
5
Re: the volume of the solid generated by the revolution

i think shell method is best
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Since more than 24 hours has gone by since the last response by the OP, I am going to post the solution while it is still on my mind.

First, I plotted the implicit curve, letting $a=1$:

solidrev1.jpg

Now, it is fairly easy to see that the domain of the implicit function is $[0,2a)$, if we write it in the form:

\(\displaystyle y^2=\frac{x^3}{2a-x}\ge0\)

We also see that the axis of rotation, i.e., the asymptote is the line $x=2a$.

We should observe that the shell method is more practical since we can solve for $y$ but not for $x$, at least not easily. Also, the disk method would generate an improper integral, so we would need to determine if it converges. So, let's proceed with the shell method. The volume of an arbitrary shell is:

\(\displaystyle dV=2\pi rh\,dx\)

where:

\(\displaystyle r=2a-x\)

\(\displaystyle h=2\sqrt{\frac{x^3}{2a-x}}\)

Since $x$ is non-negative in the domain, we may write:

\(\displaystyle h=2x\sqrt{\frac{x}{2a-x}}\)

and so we find:

\(\displaystyle dV=4\pi(2a-x)x\sqrt{\frac{x}{2a-x}}\,dx=4\pi x\sqrt{x(2a-x)}\,dx\)

Thus, summation of the shells gives us the volume:

\(\displaystyle V=4\pi\int_0^{2a} x\sqrt{x(2a-x)}\,dx\)

Completing the square under the radical, we obtain:

\(\displaystyle V=4\pi\int_0^{2a} x\sqrt{a^2-(x-a)^2}\,dx\)

Now, if we let:

\(\displaystyle x-a=a\sin(\theta)\,\therefore\,dx=a\cos(\theta)\)

we obtain:

\(\displaystyle V=4\pi a^3\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1+\sin(\theta))\cos^2(\theta)\,d\theta= 4\pi a^3\left(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2(\theta)\,d \theta+ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2(\theta)\sin(\theta)\,d\theta \right)\)

For the first integral, using a double-angle identity for cosine, we may write:

\(\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2(\theta)\,d\theta= \frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}1+ \cos(2\theta)\,d\theta= \frac{1}{2}\left([\theta]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}+ \frac{1}{2}[\sin(2\theta)]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \right)= \frac{\pi}{2}\)

For the second integral, using the odd function rule, we may write:

\(\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2(\theta)\sin(\theta)\,d\theta=0\)

And so we have:

\(\displaystyle V=4\pi a^3\left(\frac{\pi}{2} \right)=2\pi^2a^3\)
 

ksananthu

New member
Jul 14, 2013
5
Thank you.
i tried same way and I got the answer !