Question

Determine the volume of the torgus generated by

the revolving circular region enclosed by

(x − 3)²+ y² = 1 about the y-axis.

Answer 1

Given circle is (x-3)2 + y2 = 1

Centre of circle is (3,0) and radius is 1.

We know that Volume, V = $\pi$ $\int$_a^b (R² - r²) dy

Here a = -1 and b = 1

R = 3+(1-y²)^(1/2) and r = 3-(1-y²)^(1/2)

Now V =$\Pi$ $\int$_-1¹ [9+(1-y²)+6(1-y²)^(1/2)]-[9+(1-y²)-6(1-y²)^(1/2)]

V = 12 $\Pi$ $\int$_-1¹ (1-y²)^(1/2) dy

We know that$\int$(a²-x²)dx=[(1/2)x(a²-x²)^(1/2)+(1/2)a² sin^-1(x/a) + C

So, V =12$\Pi$ [(1/2)(1-(1)²)^(1/2)+(1/2)sin^-1(1)-(1/2)(1-(-1)²)^(1/2)-(1/2)sin^-1(-1)]

V = 12$\Pi$ * $\Pi$ = 12 $\Pi$ ² Units³