Problem of a solid of revolution. When I need to use the $2\pi$ in the shell method. Exactly why is used that $2\pi$

Question

The semicircular region limited by $x=\sqrt{4-y^2}$ and the Y-axis and revolves around the line $x =-1$. generated and calculate its volume.

First: the function; I can't make the graph.

So, I´m not pretty good at this still, but when the teacher teach us the subject in the shell method use a $2\pi$ at the beginning of the formula when we evaluate in x, but I don´t really get it how to use it, and I think in this problem this can be really helpful if you guys can explain to me.

Thanks in advance.

Answer 1

If you can't make the plot, you will not succeed. Just take a bunch of values of $y$, calculate the value of $x$ that goes with them, and plot the points. You should get a semicircle of radius $2$.

Intuitively, each shell is a thin cylinder. It is generated by revolving a rectangle of the region around your axis. In this case, the axis of revolution is parallel to the $y$ axis. The rectangle is from $x$ to $x+dx$ and from $y=0$ to $y=\sqrt {4-x^2}$. When you revolve this around $x-1$ the radius is $x+1$, so the circumference is $2\pi(x+1)$ The height is $\sqrt {4-x^2}$ and the thickness is $dx$, so the total volume is $2\pi(x+1)\sqrt {4-x^2}dx$. Now integrate this over the range of $x$ and you are done.

Answer 2

A good way to think about revolutions is to find the volume of only one cross sectional slice of the solid in terms of $x$ and then integrate from there. First to get the the radius of our cross sectional slice, we need to solve for $y$ because we are revolving around $x=-1$ and our differential will be $dx$.

$$y=\sqrt{4-x^2}$$

The radius, we'll call it $r$, is given by $\sqrt{4-x^2}+1$ and we know the area of a circle is $\pi r^2$. So to get the volume of our slice $v$ we just multiply $\pi r^2$ by $dx$.

$$v=\pi(\sqrt{4-x^2}+1)^2dx$$

Now to get the total volume $V$, we sum up all of the slices.

$$V=\int_0^1 \pi(\sqrt{4-x^2}+1)^2dx$$