#### To determine

**To find:**

The volume of a frustrum of a right circular cone with height h, lower base radius R, and top radius r.

#### Answer

πhR2+Rr+r23

#### Explanation

**1) Concept:**

i. Use the definition of volume

ii. A frustum of a cone is the part of the cone that remains after the top of the cone is cut-off parallel to the base of the cone.

**2) Definition of volume:**

Let S be a solid that lies between x=a and x=b. If the cross sectional area of S in the plane Px, through x and perpendicular to the x-axis, is A(x), where A is a continuous function, then the volume of S is

V=limn→∞∑i=1nAxi*∆x=∫abAxdx

**3) Given:**

Height is h, lower base radius is R, top radius is r.

**4) Calculations:**

Since the frustum has rotational symmetry, the above region shall produce the frustum upon rotation about x-axis. Consider a representative rectangle in it .

The slope of the line with end points (h, r) and (0, R) is r-Rh-0=r-Rh, and the y-intercept is R.

Using the slope intercept form, the equation of the line that represents the side of the frustum is y=r-Rhx+R.

The figure represents the disk (washer with no hole) that is generated from revolving the representative rectangle about x-axis.

The radius is,

r'=r-Rhx+R

So, the volume of the solid of revolution that is the volume of the frustum is,

Vfrustum=∫abAxdx =∫x=0x=h πr-Rhx+R2 dx=∫x=0x=h πr-R2hx2+2R·r-Rhx+R2 dx

After integrating,

Vfrustum=πr-R2h2·x33 +2R·r-Rh·x22+R2xh0

By applying the Fundamental Theorem of Calculus,

πr-R2h2·h33 +2R·r-Rh·h22+R2h-πr-R2h2·033 +2R·r-Rh·022+R20

Vfrustum=πr-R2h2·h33 +2R·r-Rh·h22+R2h

After simplifying,

Vfrustum=πhr-R23 +Rh·r-R+R2h

Vfrustum=πhr-R2 +3Rh·r-R+3R2h3

After expanding the term inside bracket,

Vfrustum=πhr2-2rR+R2+3rRh-3R2h+3R2h3

After cancelling the common terms and simplifying the equation,

Vfrustum=πr2h+rRh+R2h3= πh3(r2+rR+R2)

**Conclusion:**

The volume of a frustum of a right circular cone with top radius r, bottom radius R and height h is

Vfrustum=πh3(r2+rR+R2)