Calculus volume question

Question

The region enclosed by the circle defined by the equation $(x-2)^2+y^2 = 4$ is rotated about the y-axis to generate a solid of revolution. Find the volume of the solid.

The suggest answer is $\int_0^4$ $2{\pi}x2{\sqrt{(4-(x-2)^2)}}$ and I want to know why it is $2{\sqrt{(4-(x-2)^2)}}$ instead of ${\sqrt{(4-(x-2)^2)}}$, where the 2 came from? Thanks.

Answer 1

The area will be given by

$$A=\int_0^42\pi rh\,dx$$

where $r=x$ and $h=2y=2\sqrt{4-(x-2)^2}$

thus

$$A=\int_0^44\pi x\sqrt{4-(x-2)^2}\,dx$$

Answer 2

The short answer to you question is that the integral that you show only accounts for that part of the positive values of $y$, in other words, the semicircle, rather than the circle. Another way to look at this is in terms of Pappus's $2^{nd}$ Centroid theorem: the volume of a planar area of revolution is the product of the area $A$ and the length of the path traced by its centroid $R$, i.e., $2πR$. Then $V=2πRA$. Clearly, $R=2$, $A=\pi r^2=4\pi$ so that $V=16\pi^2$. This is the result you should get.

Another way to look at this is to cite the volume of a torus, namely $V=2\pi^2Rr^2$, where $R$ is distance to the center of the circle and $r$ is the diameter of the circle.

Answer 3

The red circle represents $(x-2)^2+y^2=4$. But do you see that blue "line". Because I couldn't draw an infinitesimal rectangle with the programs I have, just think of that blue object as a infinitesimal rectangle. In general, this rectangle is located a $x$ coordinate of $x$. As such, this rectangle is one with width $dx$ and height:

$$\sqrt{4-(x-2)^2}-\left(-\sqrt{4-(x-2)^2}\right)=2\sqrt{4-(x-2)^2}$$

I hope this clears doubt.

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Intuitively: We rotate this rectangle around the $y$ axis and it creates a cylindrical shape, we unwind that cylindrical like shape and find it to be pretty much a rectangle prism, one of dimensions $2\pi x$ by $2\sqrt{4-(x-2)^2}$ by $dx$. Hence the volume of just this shape obtained by rotating this infinitesimal rectangle is $2\pi x(2\sqrt{4-(x-2)^2}) dx$. Hence the total volume of the surface of revolution is $\int_{0}^{4} 2\pi x(2\sqrt{4-(x-2)^2}) dx$.

Answer 4

Because the circle is symmetric. When you use shell method to calculate the volume, you need to consider the area below x axis,