#### To determine

**To find:** The formula for the approximate volume of the thin cylindrical shell with height *h*, inner radius *r* and thickness Δr.

#### Answer

The formula for approximate value of the cylindrical shell is ΔV≈2πrhΔr.

#### Explanation

**Calculation:**

The volume of the cylindrical shell with *r* is the radius and *h* is height is V=πr2h.

Here, the value *h* is constant and the value *r* is changeable variable.

The differential is dV=f′(r)dr.

The derivative of the function f(r)=πr2h is computed as follows.

f′(r)=ddr(πr2h)=πhddr(r2)=2πrh

Substitute f′(r)=2πrh in dV=f′(r)dr.

dV=2πrhdr

The thickness of the cylindrical shell is Δr. That is, dr=Δr.

ΔV≈2πrhΔr

Therefore, the formula for approximate value of the cylindrical shell is ΔV≈2πrhΔr.

#### To determine

**To find:** The error involved in using the formula from part (a).

#### Answer

The error is πh(Δr)2.

#### Explanation

**Calculation:**

The volume of the cylindrical shell with *r* is the radius and *h* is height is V1=πr2h.

The volume of the cylindrical shell with radius is r+Δr and the height *h* is V2=π(r+Δr)2h^{.}

The volume of the thin cylindrical shell with height *h*, inner radius *r* and thickness Δr is computed as,

δV=V2−V1=π(r+Δr)2h−πr2h=πh(r2+2rΔr+(Δr)2−r2)=2πrh(Δr)+πh((Δr)2)

The error δV−ΔV is computed as follows.

δV−ΔV=2πrh(Δr)+πh((Δr)2)−2πrhΔr (ΔV≈2πrhΔr)=πh(Δr)2

Therefore, the error is πh(Δr)2.