Use differentials to find the approximate amount of copper in the four sides and bottom of a rectangular copper tank that is $6$ feet long, $4$ feet wide and $3$ feet deep inside, if the sheet-copper is $\dfrac14$ inch thick.
I know that $v(x,y,z)=xyz$, and $$\mathrm{d}v=Vx\,\mathrm{d}x + Vy\,\mathrm{d}y +Vz\,\mathrm{d}z$$ which is equal to $yz\,\mathrm{d}x + xz\,\mathrm{d}y + xy\,\mathrm{d}z$.
I'm not sure how to proceed from here.
On
I think you can approach this by treating the tank as two separate boxes with different lengths ($l$), widths ($w$), and depths ($d$):
Box $1$: $l\times w\times d=6\times 4\times 3$ (find volume $V_1$ by integration)
Box $2$: $(l-0.5)\times (w-0.5)\times (3-0.25)=(6-0.5)\times (4-0.5)\times (3-0.25)$ (find volume $V_2$ by integration)
Thus, copper in the tank: $V=V_1-V_2$
You are supposed to think of the $1/4$ inch sheet as a thin one. In that case you can multiply the total area of the five sides of the tank by the thickness to get the volume. This is not exact because the copper in the corners is either counted twice if the dimensions are outside or not at all if the dimensions are inside, but in this context it is the square of a small quantity and considered negligible.