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The volume of the solid generated by the revolution
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Re: the volume of the solid generated by the revolution
So do I...can you explain why?
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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 $a=1$:
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:
\(\displaystyle V=4\pi a^3\left(\frac{\pi}{2} \right)=2\pi^2a^3\)
First, I plotted the implicit curve, letting $a=1$:
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:
\(\displaystyle V=4\pi a^3\left(\frac{\pi}{2} \right)=2\pi^2a^3\)
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