You are watching: A frustum of a pyramid with square base of side b, square top of side a, and height h

A volume element is

$$dV = A(y) dy$$

where the sides in the square cross-sectional area $A(y)$ behaves linearly through height:

$$A(y) = left< b-fracyh (b-a) ight >^2 = b^2-2 b (b-a)fracyh +frac(b-a)^2 h^2 y^2$$

so the integral is

$$V = int_0^h dy , A(y) = b^2 h - b (b-a) h +frac13 (b-a)^2 h$$

or, simplifying,

$$V = frac13fracb^3-a^3b-a h $$

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The rule behind using single integrals to calculate volumes is this: One possible an approach of looking in ~ what the integral go in the usual feeling is that it switch calculations that area right into calculations that "weighted length". That is, you have the right to assign "weights" come every point on the $x$-line and then "add castle up" to acquire the area. The weights you desire are exactly the

*height*, the is, the normal size in the perpendicular direction.

It turns out the what"s important in the calculation the area is just that the weights measure the ingredient in the perpendicular direction. So by analogy we can think of the integral together calculating the volume of an item in this way: attract a line with the object, and assign weights come each allude on that line offered by the measure up in the perpendicular directions (in the cross-section). This time, there are numerous lines perpendicular, so the measure up is no longer length, yet is rather *area*.

So now rather of $A=int h(x),dx$ where $A$ is area and $h$ is height, we have actually $V=int A(x),dx$, wherein $V$ is volume and $A$ is cross-sectional area. (Addressing Makholm"s comment, we can use this principle once $A(x)=pi r(x)^2$ to recuperate the disk an approach formula which friend are familiar with.)

To be a bit an ext concrete about this, we have the right to do a silly example: calculating the volume of a cube (with side length $s$). Of course we already know it"s $s^3$, yet let"s watch if we can see exactly how to carry out it v integrals. Put your $x$-axis on among its sides, and integrating along it. At every point, the cross-section is a square with side-length $s$, for this reason its cross-sectional area is $s^2$. Because that convenience, placed one edge of the sheet on the $x$-axis at $x=0$, and one at $x=s$. Then we have the right to integrate

$$V = int_edge A(x),dx = int_0^s A(x),dx = int_0^s s^2 dx.$$$$V = s^2int_0^s 1,dx = s^2xBig|_0^s = s^3-0=s^3. $$

And the course that is the volume of a cube, favor we currently knew. Hope this is rather convincing :)

You can see some major themes here. The most necessary thing is the right an option of line. If we had chosen the diagonal line of the cube, we absolutely could perform the calculation, yet it"s significantly messier. So friend should try to make use of some symmetry when picking the heat so the the area function will be nice to play approximately with. We also have the flexibility of wherein to ar the origin; usually putting it at the end of the number will do calculations easier, back the facility is one more place the is often effective as well.