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?
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 $$V=\pi\int_a^b [f(x)]² dx-\pi\int_a^b [g(x)]² dx$$