Question

Find the volume of the solid generated by revolving the circle x^2 + y^2 = 4 about the line x = 3. Use the cylindrical shell method.


Note: With sketch please

Answer 1

Consider $x^2+y^2=4, y\geq0.$ Then $y=\sqrt{4-x^2}$

Figure shows the region and a cylindrical shell formed by rotation about the line $x=3.$ It has radius $3-x,$ circumference $2\pi(3-x),$ and height $\sqrt{4-x^2}.$

The volume of the given solid is

$$V_1=\displaystyle\int_{-2}^22\pi(3-x)\sqrt{4-x^2}dx$$

$x=2\sin{t},\ -\pi/2\leq t\leq \pi/2$

$\ dx=2\cos{t}dt$

$\sqrt{4-x^2}=2\cos{t}$

$$\int2\pi(3-x)\sqrt{4-x^2}dx=$$

$$=2\pi \int(3-2\sin{t})(2\cos{t})(2\cos{t})dt=$$

$$=24\pi \int\cos^2{t}dt-16\pi\int\sin{t}\cos^2{t}dt=$$

$$=12\pi \int(1+\cos{2t})dt+16\pi ({\cos^3{t} \over 3})=$$

$$=12\pi t+6\pi \sin{2t}+{16\pi \over 3}\cos^3{t}+C=$$

$$=12\pi\arcsin{({x \over 2})}+3\pi x\sqrt{4-x^2}+{2\pi \over 3}(4-x^2)^{3/2}+C$$

$$V_1=\pi\bigg[12\arcsin{({x \over 2})}+3x\sqrt{4-x^2}+{2\over 3}(4-x^2)^{3/2}\bigg]\begin{matrix} 2 \\ -2 \end{matrix}=$$

$$=\pi(12({\pi \over 2})+0+0-12(-{\pi \over 2})-0-0)=12\pi^2 (units^3)$$

$$V=2V_1=2(12\pi^2)=24\pi^2(units^3)$$

The volume of the solid generated by revolving the circle $x^2+y^2=4$ about the line $x=3$ is $24\pi^2$ cubic units.