# Why doesn't the second method work?

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?

• Surely what you're calculating is a volume, not area. – J.G. Jan 26 at 17:09
• @J.G. i meant volumes, sorry – 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 $$\pi[f(x)]²dx$$.
$$V=\pi\int_a^b [f(x)]² dx$$.
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 $$V=\pi\int_a^b [f(x)]² dx-\pi\int_a^b [g(x)]² dx$$