I have been trying to complete this question:

The region R, bounded by the curve with equation $y=\sin(x)$, $0\leqslant x \leqslant \pi$ and the line with equation $y = \dfrac{1}{\sqrt2}$

The region R is rotated through $2\pi$ radians about the line $y = \dfrac{1}{\sqrt2}$

Show that the solid of revolution formed has area $\dfrac{\pi}{2}(\pi-3)$

I have tried two methods, one of which has worked, and the other has not, but I don't understand why the second method didn't work

The first method was to do the integral $$\pi\int_{\dfrac{\pi}{4}}^{\dfrac{3\pi}{4}} (\sin^2x-\dfrac{1}{\sqrt2})^2 dx$$, which worked fine.

The second method was to do the integral $$\pi\int_{\dfrac{\pi}{4}}^{\dfrac{3\pi}{4}} \sin^2x dx - \pi\int_{\dfrac{\pi}{4}}^{\dfrac{3\pi}{4}} \dfrac{1}{2} dx$$, which would find the volume from rotating $y=\sin x$, and subtract the volume from rotating $y=\dfrac{1}{\sqrt2}$, which I thought would find the correct value, but it didn't.

I can see that the integrals are different, because the expansion of $(\sin^2x-\dfrac{1}{\sqrt2})^2$ $\ne$ $\sin^2x-\dfrac{1}{\sqrt2}$, but I am not getting why the second method doesn't work, as it looks like it should, graphically.

My question is why doesn't the second method work, but the first does?

  • $\begingroup$ Surely what you're calculating is a volume, not area. $\endgroup$ – J.G. Jan 26 at 17:09
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    $\begingroup$ @J.G. i meant volumes, sorry $\endgroup$ – Joshua Peacham Jan 26 at 17:15

Where did you find this question? The second integral seems to me the correct way to evaluate the volume. When calculating the volum of a revolution solid using this method, what are you doing in fact is "adding" the areas of infinite cilinders with very small heights, each one with radius $f(x)$ and "height" $dx$. The "volume" of each cilinder is $\pi[f(x)]²dx$.

So the volume of the solid is

$$V=\pi\int_a^b [f(x)]² dx$$.

(You can make this rigorous using Riemann sums, but I think it's unnecessary here.)

In your case, you're not adding cylinders, you are adding "cylinders" with a hole, so the "volume" of each cylinder is $[\pi f(x)²-\pi g(x)²]dx$. "Adding it up", it would result in $$V=\pi\int_a^b [f(x)]² dx-\pi\int_a^b [g(x)]² dx$$

  • $\begingroup$ It's from my school textbook, I thought the second method should work as it finds the total volume from rotating y=sinx then subtracting the cylinder that we don't want, as it is the part outside the region R $\endgroup$ – Joshua Peacham Jan 26 at 17:14

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