Question

Figure shows a semicircle that is the graph of the equation $y=\sqrt{6x-x^2}$. If the semicircle is rotated ${360}^{\circ }$ about the $x-$axis, calculate the volume of the sphere that is created. $$$$
 

A.$6\pi$ $$$$
B. $12\pi$ $$$$
C. $18\pi$ $$$$
D. $24\pi$ $$$$
E.$36\pi$ $$$$

Answer 1

Hint: We denote the other point of intersection except origin as A. We find the coordinate of A by finding the solutions of $y=\sqrt{6x-x^2}=0$. We find OA as the diameter of the rotating sphere and find the volume of sphere $V={\displaystyle \frac{4}{3}}\pi r^3$ where $r$ is the radius of the sphere.

Complete step-by-step solution:
We see in the figure that a semicircle is present whose one end is at the origin is defined by the equation
$$y=\sqrt{6x-x^2}........\left(1\right)$$
We denote the other end of the semicircle lying on the positive $x-$axis as A.
 

Since the equation of $x-$axis is $y=0$ . So we can find the coordinates of A since the semicircle A lies on the $x-$axis when $y=\sqrt{6x-x^2}=0$. So we have;
$$\sqrt{6x-x^2}=0$$
We square both sides of above equation to have;
$$\Rightarrow 6x-x^2=0$$
We factorize by taking $x$ common in the above step and have;
$$\Rightarrow x(6-x)=0$$
Since the product of the two factors is zero one of them must be zero. So we have;
$$\begin{array}{rl}& \Rightarrow x=0\text{ or }6-x=0\\ & \Rightarrow x=0\text{ or }x=6\end{array}$$
So we have two roots of the equation $x=0,6$. So when $x=0,y=0$ we have the coordinate of the origin $O(0,0)$ and when $x=6,y=0$ we have a coordinate of $A(6,0)$. So the diameter of the circle is $OA=6$ units. $$$$
We are given the question that the semicircle is rotated ${360}^{\circ }$ about the $x-$axis and creates a sphere. So the diameter of the sphere is $OA=6$ units. So the radius $r$ of the sphere is $r={\displaystyle \frac{OA}{2}}={\displaystyle \frac{6}{2}}=3$ units. Then volume $V$ of the sphere in cubic units is
$$V={\displaystyle \frac{4}{3}}\pi r^3={\displaystyle \frac{4}{3}}\pi {\left(3\right)}^3={\displaystyle \frac{4}{3}}\pi \left(27\right)=4\times 9\pi =36\pi$$
So the correct option is E.

Note: We note that the question presumes that the semicircle is not displaced from the original end points O and A when it's rotated about the $x-$axis. We can alternatively find the radius by squaring the given equation $y=\sqrt{6x-x^2}$ and then comparing it with general equation of circle ${(x-a)}^2+{(y-b)}^2=r^2$ where $r$the radius is and $(a,b)$ is the centre of the circle. The other semi-circle fourth quadrant will have the equation$y=-\sqrt{6x-x^2}$.