By the Pappus Theorem, the volume of the solid formed by revolving the region enclosed by the circle $( x - 3 ) ^ { 2 } + y ^ { 2 } = 1$ about the y-axis is

A)1
B) $\frac { 2 } { 5 }$
C) $\frac { 6 } { 5 }$
D) $\frac { 3 } { 4 }$
E) $6 \pi ^ { 2 }$

Question

By the Pappus Theorem, the volume of the solid formed by revolving the region enclosed by the circle $( x - 3 ) ^ { 2 } + y ^ { 2 } = 1$ about the y-axis is

A)1
B) $\frac { 2 } { 5 }$
C) $\frac { 6 } { 5 }$
D) $\frac { 3 } { 4 }$
E) $6 \pi ^ { 2 }$

Quizplus · Accepted Answer

Pappus's Theorem states that the volume of a solid of revolution generated by revolving a plane figure about an external axis is equal to the product of the area of the figure and the distance traveled by its centroid. The given circle has a radius of 1 and is centered at (3,0), so its area is $\pi r^2 = \pi(1)^2 = \pi$. The centroid of a circle is its center, so the distance traveled by the centroid during the revolution is the circumference of a circle with radius equal to the distance from the y-axis to the center of the given circle, which is 3. Thus, the distance traveled by the centroid is $2\pi(3) = 6\pi$. Therefore, the volume is $\pi \times 6\pi = 6\pi^2$.

By the Pappus Theorem, the Volume of the Solid Formed (x−3)2+y2=1( x - 3 ) ^ { 2 } + y ^ { 2 } = 1(x−3)2+y2=1

By the Pappus Theorem, the Volume of the Solid Formed $( x - 3 ) ^ { 2 } + y ^ { 2 } = 1$