volume of revolution

Harry_the_cat

Harry_the_cat

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Mar 16, 2016
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Would someone mind checking my solution please?
I am trying to find the volume of revolution generated by rotating the minor segment bounded by the line y=x and the circle with centre (1,2) and radius 1, about the line y=x.
What I did was rotate the whole thing 45 degrees clockwise about the origin and reduced the problem to a rotation about the x-axis. The new circle had centre \(\displaystyle (\frac{3}{\sqrt2}, \frac{1}{\sqrt2}) \)and x-intercepts \(\displaystyle \sqrt2\) and \(\displaystyle 2\sqrt2\).

The solution I got is 0.669 cubic units. Anyone like to check my give it a go and see if you get the same result?

Or is there another way to do it?
 
By inspection, I find that the volume \(V\) may be written:

[MATH]V=2\pi\int_{0}^{\frac{1}{\sqrt{2}}} \left(\sqrt{1-x^2}-\frac{1}{\sqrt{2}}\right)^2\,dx=\frac{\pi(10-3\pi)}{6\sqrt{2}}\approx0.21297[/MATH]
If I check that using the formula for rotation about an oblique axis, I find:

[MATH]V=\frac{\pi}{2^{\frac{3}{2}}}\int_{1}^{2}\left(-\sqrt{1-(x-1)^2}+2-x\right)^2\left(1+\frac{x-1}{\sqrt{1-(x-1)^2}}\right)\,dx=\frac{\pi(10-3\pi)}{6\sqrt{2}}\approx0.21297\quad\checkmark[/MATH]
 
MarkFL said:
By inspection, I find that the volume \(V\) may be written:

[MATH]V=2\pi\int_{0}^{\frac{1}{\sqrt{2}}} \left(\sqrt{1-x^2}-\frac{1}{\sqrt{2}}\right)^2\,dx=\frac{\pi(10-3\pi)}{6\sqrt{2}}\approx0.21297[/MATH]
If I check that using the formula for rotation about an oblique axis, I find:

[MATH]V=\frac{\pi}{2^{\frac{3}{2}}}\int_{1}^{2}\left(-\sqrt{1-(x-1)^2}+2-x\right)^2\left(1+\frac{x-1}{\sqrt{1-(x-1)^2}}\right)\,dx=\frac{\pi(10-3\pi)}{6\sqrt{2}}\approx0.21297\quad\checkmark[/MATH]
Click to expand...
Thankyou so much Mark. That confirms my answer. Really appreciate you taking the time to do the problem. Thanks again.
 
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