#### To determine

**To find:**

The volume of the solid torus.

#### Answer

2π2Rr2 unit3

#### Explanation

**1) Concept:**

i. If x is the radius of the typical shell, then the c ircumference =2πx and the height is y

ii. By the shell method, the volume of the solid by rotating the region under the curve y=f(x) about y- axis from a to b is

V= ∫ab2π x f(x)dx

where 0≤a≤b

Integral of an odd function is zero

The area of the semicircle of the radius r is πr22.

**2) Given:**

Solid torus

**3) Calculation:**

The torus is shown in figure.

The torus consists of the circle of the radius r centred at (R, 0), rotated along the y axis.

The equation of the circle with the points (x1, y1) is

x-x12+y-y12=r2

Therefore the equation of the circle with centre (R, 0) is

x-R2+y2=r2

The volume can be found out by rotating the circle of the radius r along the y axis. So, use the shell method to find the volume of the torus.

As the region is bounded by x-R2+y2=r2, that is, y2=r2-x-R2

⇒y=± r2-x-R2

Let the given region rotated about the y- axis.

Using the shell method, find the typical approximating shell with radius x.

Therefore, the circumference is 2πx and the height is

y=r2-x-R2-- r2-x-R2=2r2-x-R2orus h and e cone in figure.in figure.ral of odd function.o

So, the total volume is

V= ∫ab2πx 2r2-(x-R)2dx

V=4π∫R-rR+r(x)r2-(x-R)212dx

Let x-R=t⇒dx=dt

x=R+r⇒t=r and x=R-r⇒t=-r

V= 4π∫-rr(R+t)r2-t212dt

V= 4πR∫-rrr2-t212dt+ ∫-rrtr2-t212dt

The first integral in the parenthesis is the area of the semicircle of the radius r and is equal to πr22

And the second integral is zero because it is the integral of an odd function.

V= 4πRπr22+0

V= 4πRπr22

V=2π2Rr2

Therefore, the volume of the sphere of the radius rV= 2π2Rr2unit3

**Conclusion:**

The volume of the sphere of the radius r is V= 2π2Rr2unit3