Find the volume of the solid obtained by rotating the region of $(x-1)^2+(y-1)^2 \le 1$ about the $y=\frac12$

Question

Find the volume of the solid obtained by rotating the region of $(x-1)^2+(y-1)^2 \le1$ about the $y=\frac12$

This is the figure that you get when you try to show on the Oxy system. I'm struggling with how to use the formula of the volume in this case.

Is this right what I did? I don't know really what shape will I get $$\mathsf{V}=\pi\int_0^{1-\frac{\sqrt3}{2}}\left(1+\sqrt{1-(x-1)^2}-\frac12\right)^2-\left(1-\sqrt{1-(x-1)^2}-\frac12\right)^2dx+\pi\int_{1-\frac{\sqrt3}{2}}^{1+\frac{\sqrt3}{2}}\left(1+\sqrt{1-(x-1)^2}-\frac12\right)^2dx+\pi\int_{1-\frac{\sqrt3}{2}}^{1+\frac{\sqrt3}{2}}\left(\frac12-(1-\sqrt{1-(x-1)^2})\right)^2dx+\pi\int_{1-\frac{\sqrt3}{2}}^2\left(1+\sqrt{1-(x-1)^2}-\frac12\right)^2-\left(1-\sqrt{1-(x-1)^2}-\frac12\right)^2dx$$

Answer 1

Calculates the volume of the body formed by the rotation of the segment above line $y=\dfrac12.$

The segment is symmetric about the line $x=1,$ so the volume can be calculated for left part factor $2$.

The integral is splitted to the fragments $x\in\left(0,1-\frac{\sqrt3}3\right)$ and $x\in\left(1-\frac{\sqrt3}3\right).$

The volume equals to $$V=2\pi\left(\int\limits_{0}^{1-\frac{\sqrt3}2}\left(\left(1+\sqrt{1-(x-1)^2}-\frac12\right)^2-\left(1-\sqrt{1-(x-1)^2}-\frac12\right)^2\right)\,\mathrm dx + \int\limits_{1-\frac{\sqrt3}2}^1\left(1+\sqrt{1-(x-1)^2}-\frac12\right)^2\,\mathrm dx\right),$$ or $$V=2\pi\left(\int\limits_{0}^1\left(1+\sqrt{1-(x-1)^2}-\frac12\right)^2\,\mathrm dx -\int_0^{1-\frac{\sqrt3}2}\left(1-\sqrt{1-(x-1)^2}-\frac12\right)^2\,\mathrm dx \right).$$ Therefore, the inner part subtracts from the outer one.

Answer 2

The way you have stated this problem, the volume is clearly 0 because the rotated figure is the surface of a three dimensional solid, not the solid itself. I suspect you meant $(x- 1)^2+ (y- 1)^2\le 1$. Also note that only the part with $y\ge 1/2$ counts in calculating the rotated area. $y< 1/2$ is "inside" the other.

Now, what "formula of the volume" are you talking about? With a rotation problem you probably want to use the "disk method", the "washer method" or the "shell method".

When y= 1/2, $(x- 1)^2+ 1/4= 1$ so $x= 1\pm \frac{\sqrt{3}}{2}$. To cover the figure using the "disk" and "washer" methods, you need to divide the integration into three parts: x from 0 to $1- \frac{\sqrt{3}}{2}$, x from $1- \frac{\sqrt{3}}{2}$ to $1+ \frac{\sqrt{3}}{2}$, and x from $1+ \frac{\sqrt{3}}{2}$ to 2. On the first and third intervals, the "washer" has outer radius $y= 1+ \sqrt{1- (x-1)^2}$ and inner radius $y= 1- \sqrt{1- (x-1)^2}$. On the third interval, the "disk" has radius $1+ \sqrt{1- (x-1)^2}- \frac{1}{2}$.

Using the "shell method" you would integrate over y from 1/2 to 2 with each shell having radius y- 1/2 and length $(1+ \sqrt{1- (y-1)^2}- (1- \sqrt{1- (y-1)^2})= 2\sqrt{1- (y- 1)^2}$

Answer 3

Let's first translate $\left(1, \frac{1}{2} \right)$ to the origin and revolve around the $x$-axis (since there are standard formulae for this), so we're now considering $$x^2 + \left(y-\frac{1}{2}\right)^2 \leq 1$$ Observe that the volume formed by rotating the segment in the $y \leq 0$ region is contained within that of the volume from the $y \geq 0$ region, so to calculate the volume of the solid we consider only the $y \geq 0$ region:

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The $x$ axis intersects the circle at $y = 0 \implies x^2 = \frac{3}{4}$, and since the centre of the circle is at $\left(0, \frac{1}{2}\right)$, we have that the point of intersection $\left(0, \frac{\sqrt{3}}{2} \right)$ makes an angle $\alpha$ with the line $y = \frac{1}{2}$ (i.e. with the centre of the circle) where $\alpha$ is given by $$\alpha = -\sin^{-1}\frac{1}{2} = -\frac{\pi}{6}$$ and the other point of intersection makes an angle of $\pi + |\alpha| = \frac{7\pi}{6}$. This means we can parametrise the curve shown above by $$(x, y) = \left(\cos\theta, \, \frac{1}{2} + \sin\theta \right), \quad -\frac{\pi}{6} \leq \theta \leq \frac{7\pi}{6}$$ Now, the volume generated by revolving this curve around the $x$-axis is given by (for example, see this Wikipedia page) $$V = \pi \int_{\theta_1}^{\theta_2} y^2 \frac{dx}{d\theta} \; d\theta = \pi \int_{7\pi/6}^{-\pi/6} \left(\frac{1}{2} + \sin \theta\right)^2 \cdot -\sin \theta \; d\theta$$ where we've chosen the limits so that we go from negative to positive $x$ (we could just take the absolute value afterwards if this came out negative). Expanding this out, we get $$V = \pi \int_{-\pi/6}^{7\pi/6} \left( \sin^3\theta + \sin^2\theta + \frac{\sin \theta}{4} \right) \; d\theta$$ Integrating the various powers of $\sin$ that appear is best done by using appropriate multi-angle identities. Making use of these, the integrand becomes $$\frac{3 \sin \theta - \sin3 \theta}{4} + \frac{1 - \cos2 \theta}{2} + \frac{\sin \theta}{4} = \frac{1}{2} + \sin \theta - \frac{\sin 3 \theta}{4} - \frac{\cos 2 \theta}{2}$$ and hence $$V = \pi \left[ \frac{\theta}{2} - \cos \theta + \frac{\cos 3 \theta}{12} - \frac{\sin 2 \theta}{4} \right]_{-\pi/6}^{7\pi/6} = \frac{3 \sqrt 3}{4} + \frac{2 \pi}{3}$$ is the required volume.