Question

Use differentials to estimate the amount of material in a closed cylindrical can that is 20 cm high and 8 cm in diameter if the metal in the top and bottom is 0.1 cm thick, and the metal in the sides is 0.1 cm thick. Note, you are approximating the volume of metal which makes up the can (i.e. melt the can into a blob and measure its volume), not the volume it encloses

Answer 1

Answer:

The volume is  $dV = 19.2 \pi \ cm^3$

Step-by-step explanation:

From the question we are told that

    The height is  h =  20 cm

    The diameter is  d = 8 cm

    The thickness of both top and bottom is  dh  =  2 * 0.1  =  0.2 m

     The  thickness of one the side is dr =  0.1 cm

The  radius is mathematically represented as

            $r = \frac{d}{2}$

substituting values

            $r = \frac{8}{2}$

           $r = 4 \ cm$

Generally the volume of a cylinder is mathematically represented as      

       $V_c = \pi r^2 h$

Now the partial differentiation with respect to h is  

       $\frac{\delta V_v}{\delta h} = \pi r^2$

Now the partial differentiation with respect to r is  

      $\frac{\delta V_v}{\delta r} = 2 \pi r h$

Now the Total differential of $V_c$ is mathematically represented as

       $dV = \frac{\delta V_c }{\delta h} * dh + \frac{\delta V_c }{\delta r} * dr$

       $dV = \pi *r^2 * dh + 2\pi r h * dr$

substituting values

      $dV = \pi (4)^2 * (0.2) + (2 * \pi (4) * 20) * 0.1$

     $dV = 19.2 \pi \ cm^3$

Answer 2

(I deleted my answer because it was incorrect)