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I've been solving problems on cylindrical shells and this is the formula the textbook has given me: $V=\int_a^b 2\pi xf(x)\:dx$, where $x$ is the radius of the shell.

I usually just figure out the upper and lower bounds if there's more than one function, use them to define $f(x)$, then figure out the integral limits, and substitute in the formula above. However, for this problem, I cannot figure out how I am supposed to apply the shells method.

To further clarify, here is the exact question with all the given data.

"Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves about the x-axis. $y=x$, $x=0$ and $x+y=2$."

PS- I have to use the cylindrical shells method. I realize this might be a very basic question, but integration isn't my strong suit and I have trouble visualizing problems like these.

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    $\begingroup$ $x=0$ is the y-axis. The other are two lines. Graph them all. y=x intersects x=0 at the origin. (solve the one unknown using the known value of x$, do the same for each pair of equations. This is early pre-calc. Given that what I describe is "no sweat" for you, please edit your post with the bounds of integration. $\endgroup$
    – amWhy
    Commented Feb 19, 2022 at 21:22

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From what amWhy said, we see that $x=0$ is the $y$-axis. So let's graph this region. We get the following where $y = 2-x$ (the same as $x + y = 2$) is the red line and the blue line is $y=x$. We see that these two points intersect at $(1,1)$ and since we are integrating with respect to $x$ we will integrate from where $x=0$ to the intersection, $x=1$. enter image description here

For finding what $f(x)$ is we will do the $top - bottom$ to see what the area it is what we are integrating. $f(x) = 2-x-x = 2-2x$.

Now we will perform the integration since we have all the pieces of the puzzle. $$\int_0^1 2\pi x(2-2x) dx = 2\pi \int_0^1 x(2-2x)dx \\ = 2\pi \int_0^1 2x - 2x^2dx = 2\pi \big[ \frac{2x^2}{2} - \frac{2x^3}{3}\big] \big|_0^1 \\ = 2\pi \big[x^2 - \frac{2}{3}x^3 \big]\big|_0^1 = 2\pi \big[1 - \frac{2}{3} \big] = 2\pi \big[\frac{1}{3} \big] \\ = \boxed{\frac{2\pi}{3}}$$

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  • $\begingroup$ Thank you so much! I have one confusion though.. when we say rotating about the x-axis, does it not mean we have to take the right-left approach rather than the top-bottom (since the horizontal axis is our axis of rotation) ? In the textbook, if we are asked to rotate about the x-axis, then they integrate with respect to y. $\endgroup$
    – rayank97
    Commented Feb 20, 2022 at 7:49

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