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Find the solid of revolution obtained by rotating the region bounded by the curves $x=y^2$ and $x=1-y^2$ about the line $x=3$.

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To solve this problem I tried using the washer method with respect to $y$ (i.e. the variable being integrated was $y$, since $dy$ represents the height of each washer.

I found the formula for the area of both functions by subtracting the second function from the first, to get $A = 2y^2-1$. I then found the area of a single washer by using the formula $A = \pi r^2$. $r$ in this case is the height of the function, i.e. the value of $y$. So I plug the function I found earlier and get $A=\pi (2y^2-1)^2$ $= \pi (4y^4 - 4y^2 + 1)$

Now I integrate this, with an upper bound of 1 and a lower bound of 0, to get an answer of $7\pi /15$

I realize I may have done this completely wrong, but I can't see specifically where I went wrong. Could you point out my misunderstanding or possibly show me a better way entirely to solve this question? Any help whatsoever is greatly appreciated!

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There are two basic approaches to this. One is to integrate with respect to $x$, the other with respect to $y$. In both cases, for each value of our chosen variable we find a relevant area, and then we integrate those areas to find the volume.

With respect to $y$: The two parabolas intersect at $y = \pm\frac{\sqrt2}2$, so these are the bounds of our integral. At any fixed $y\in (-\frac{\sqrt2}2, \frac{\sqrt2}2)$, the cross-section is an annulus ("washer"), with outer radius $3-y^2$, and inner radius $3-(1-y^2) = 2+y^2$. The area is therefore $$ A_y(y) = \pi(3-y^2)^2 - \pi(2+y^2)^2 $$ Thus our volume is given as $$ \int_{-\sqrt2/2}^{\sqrt2/2}A(y)\,dy \\ = \int_{-\sqrt2/2}^{\sqrt2/2}\pi\Big(5 - 10y^2\Big)dy= \frac{10\sqrt2\pi}{3} $$

With respect to $x$: For each value of $x$ between $0$ and $1$ (the boundaries of the region bounded by the parabolas), your solid of revolution has a corresponding cylinder "cross"-section. For $x$-values between $0$ and $\frac12$, the height of that cylinder is governed by $x = y^2$, while for $x$-values between $\frac12$ and $1$, the height is governed by $x = 1-y^2$. This makes it natural to split into two integrals. For the first half of the interval, the area is $$ A_x(x) = 2\pi(3-x)\cdot 2\sqrt x $$ while for the second half, the area is $$ A_x(x) = 2\pi(3-x)\cdot 2\sqrt{1-x} $$ For the final answer we get $$ \int_0^{1/2}4\pi(3-x)\sqrt x\,dx + \int_{1/2}^1 4\pi(3-x)\sqrt{1-x}\,dx\\ = \frac{9\sqrt2\pi}{5} + \frac{23\sqrt2\pi}{15} = \frac{10\sqrt2\pi}{3} $$


There is a caveat here which is subtle, and may seem unimportant, but in a more complicated setting it must be handled. Basically, I'm glossing over the "thickness" of these areas. This integration is, after all is said and done, about adding up very many very small volumes, not infinitely many areas.

Take the $y$ example, for instance. What we actually have are not annuli, but (very thin) so-called "hollow cylinders" (think an actual, physical washer with thickness, or a coin with a hole in it). For a very narrow $y$-interval, say $y$ from $-0.3$ to $-0.29999$, the corresponding part of the volume has some thickness to it. And the way we've set this up, that thickness is exactly the same as the difference in $y$-value from $-0.3$ to $-0.29999$.

Had I made these washers slanted instead of perpendicular to the $y$-axis (which would be a natural thing to try if, say, the rotational axis had been at an angle), then that is something I would've had to take into account, as the washers would've been thinner than the difference in $y$-value may indicate.

A real-world example of this effect can be seen when you slice a cucumber diagonally. The thickness of a slice as measured along the length of the cucumber (the width of the peel at the narrow ends of the slice) is larger than the actual thickness of the slice.

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  • $\begingroup$ (+1) Nice and neat explanation. $\endgroup$
    – Robert Z
    Mar 2, 2019 at 14:20
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No, you should not have subtracted your functions at the beginning. This problem will be much easier if you create an outer radius and an inner radius. Then you can find the volume of the outer surface and subtract it from the inner surface to give you the middle volume. To find the outer radius, you can subtract 3 from the equation of your “outer function. Same goes for your inner radius.

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Just another approach (after reading Arthur's nice answer).

The area of the cross-section to be rotated is $$\int_{x=0}^{1/2}\int_{y=-\sqrt{x}}^{\sqrt{x}}1 dx +\int_{x=1/2}^{1}\int_{y=-\sqrt{1-x}}^{\sqrt{1-x}}1 dx=4\int_{x=0}^{1/2}\sqrt{x}\,dx=\frac{2\sqrt{2}}{3}.$$ Since by symmetry the centroid of this cross-section is $G=(1/2,0)$, by Pappus's centroid theorem it follows that the volume is $$V=2\pi (3-1/2)\cdot \frac{2\sqrt{2}}{3}=\frac{10\pi\sqrt{2}}{3}.$$

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