Find the volume generated by rotating the given region about the specified line

Question

I am really unsure how to solve this problem. I was hoping someone can give me some helpful hints so I can figure it out. I also looked at some other posts but they weren't helpful

Find the volume generated by rotating the given region about the specified line. $R1$ about $x = 0$

Answer 1

No calculus is needed here other than Pappus's second centroid theorem.

First, find the centroid of the region $R_1$. That region is a triangle, so calculus is not needed: just take the average of the three vertices. (Treat the vertices as two-dimensional vectors when doing the calculation.)

Second, find the length of the path the centroid would take when the region is rotated around the given line. That is easy, since the radius of the circle that the centroid makes is just its $x$-coordinate, since you are rotating around the $y$-axis.

Third, find the area of the region $R_1$. Again, this is easy since it is a triangle.

Then Pappus's theorem says that the desired volume is

$$V=Ad$$

where $V$ is the desired volume, $A$ is the area of the rotated region, and $d$ is the length of the path of the centroid when rotated.

Finding the volumes if region $R_2$ or $R_3$ would be more difficult, since you would need calculus to find the centroid and the area. The ideas would still be the same.

Let us know if you need more details. There are other ways to solve this problem, of course, which use more calculus, but this way is probably easier for you.

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There is another way that you probably will find easier. Rotating the triangle around the $y$-axis will result in a cone. The height is the segment from the origin to $(0,1)$ and the radius of the base is the segment from $(0,1)$ to $(1,1)$. You should easily be able to use the formula for the volume of a cone to get your desired answer.

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Yet another way is the Method of disks when rotating around the $y$-axis. In this case you are integrating from the $y$-axis from the line $x=y$ from $y=0$ to $y=1$. This means the desired volume is

$$V=\pi\int_0^1 y^2\,dy$$

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Yet another way is the Method of cylindrical shells for rotating around the $y$-axis. Here you are integrating from the $y=x$ to the line $y=1$ from $x=0$ to $x=1$. This means the desired volume is

$$V=2\pi\int_0^1 x\cdot (1-x)\,dx$$

Answer 2

$\dfrac { \pi}3$, because the slope is $x$, then you square it, then you take the antiderivative of $x^2$ from $0$ to $1$ times $\pi$, and get $\dfrac {x^3}3$ from $0$ to $1$ times $\pi$, and when you plug it in, you get $\dfrac {\pi}3$.