13. Volume

a1. Volume by Slicing

Integrals can also be used to find volumes! We start with a method which slices the volume perpendicular to an axis. This time we begin with an example.

Find the volume of the solid whose base is the semicircle \(0 \le y \le \sqrt{9-x^2}\) and whose cross-sections perpendicular to the \(x\)-axis are squares. Here are pictures of the base, the solid and a cross-sectional slice through the solid at position \(x\) so we can see the square cross-section:

The plot shows a semicircle in a vertical x y plane of radius 3
with its diameter edge on the horizontal x axis.
There is a vertical line from the x axis to the circle. — The Base

The plot shows a solid which looks like a quarter of a football
with its back surface being the semicircle of the previous plot. — The Solid

The plot shows the quarter of a football from the previous plot
but it is cut off at the right end showing a slice which is a square. — The Solid Sliced at \(x\)

Caution: The base here is standing up at the back of the solid, since that is where we usually draw the \(xy\)-plane.

As with all integrals, to compute the volume, we cut the volume up into small pieces, find the volume of each piece, add them up and take the limit as the number of pieces becomes large (\(n\rightarrow\infty\)) and the size of each piece gets small (\(\Delta x\rightarrow0\)). In this case, the pieces are slices perpendicular to the \(x\)-axis, each of thickness \(\Delta x\). So the above picture should be regarded as a loaf of bread which has been sliced. Below is a picture of one slice, the slice at \(x\), and an animation showing the slices being added up to form the solid:

The plot shows an outline of the quarter football from the previous
plots plus a vertical slice which has square faces. — A Slice

The video shows an outline of the quarter football from the previous
plot which is being filled by vertical slices with square faces. — The volume is formed by adding up slices.

The volume of one slice is approximately the area of the square face times the thickness \(\Delta x\). To find the area of the square, we need to know the length of a side. But this is just the height of the semicircle at the position \(x\) of the slice. So the side is \(\displaystyle s=y=\sqrt{9-x^2}\) and the area is: \[ A(x)=s^2=\sqrt{9-x^2}^2=9-x^2 \] Hence the volume of the slice is approximately: \[ \Delta V \approx A(x)\Delta x =(9-x^2)\Delta x \] We now add up these volumes and take the limit as the number of slices becomes large. The result gives the volume as an integral which we evaluate: \[\begin{aligned} V&=\lim_{n\rightarrow\infty}\sum_{i=1}^n\Delta V =\lim_{n\rightarrow\infty}\sum_{i=1}^n(9-x^2)\,\Delta x \\ &=\int_{-3}^3 (9-x^2)\,dx =\left[9x-\dfrac{x^3}{3}\right]_{-3}^3 \\ &=(27-9)-(-27+9)=36 \end{aligned}\]

In other volume by slicing problems, the shape of the base may change and the shape of the cross-sections may change, but in all such problems, we slice the volume perpendicular to some axis, compute the cross-sectional area (This is the hard part.) and integrate over the length of the solid.

If a solid extends from \(x=a\) to \(x=b\) and the cross-sections perpendicular to the \(x\)-axis have area \(A(x)\), then the volume is: \[ V=\int_a^b A(x)\,dx \] If a solid extends from \(y=a\) to \(y=b\) and the cross-sections perpendicular to the \(y\)-axis have area \(A(y)\), then the volume is: \[ V=\int_a^b A(y)\,dy \]

In practice, you do not need to go back to first principles as in the example. Rather you will need to work to find a formula for the cross-sectional area. Draw the base; draw a cross section; find the length of the base of the cross section and find the area of the cross section.

Here is another example in which the cross-sections are perpendicular to the \(y\)-axis.

Example

Now you try some:

Find the volume of the solid whose base is the semicircle \(0 \le x \le \sqrt{9-y^2}\) and whose cross-sections perpendicular to the \(x\)-axis are rectangles whose height is half of the base.

Draw the base and one cross-section.

Here are the base with one cross-section and an accumulation of cross sections to form the final solid:

The plot shows a semicircle in a vertical x y plane of radius 3
with its diameter edge on the vertical y axis.
There is a vertical rectangle perpendicular to the x axis whose top
and bottom edges meet the circle. — The Base with
one Cross Section

The video shows the vertical semicircle of the previous plot.
The vertical rectangles are being accumulated to fill a solid. — Accumulating Cross Sections
and the Solid

\(V=36\)

See the second hint for pictures of the base with one cross-section and an accumulation of cross sections to form the final solid:

The equation of the circle is \(x^2+y^2=9\). The cross section sits on the base on the line segment from \(y_-=-\sqrt{9-x^2}\) to \(y_+=\sqrt{9-x^2}\). So its length is \(b=y_+-y_-=2\sqrt{9-x^2}\). The height is half the length, \(h=\dfrac{1}{2}b=\sqrt{9-x^2}\). So the cross-sectional area is \[ A=bh=2(9-x^2) \] and the volume is \[\begin{aligned} V&=\int_0^3 2(9-x^2)\,dx =2\left[9x-\dfrac{x^3}{3}\right]_0^3 \\[3pt] &=2(27-9)=36 \end{aligned}\]

Find the volume of the solid whose base is the circle \(x^2+y^2=4\), and whose cross-sections perpendicular to the \(x\)-axis are equilateral triangles.

Here are the base and a crosssection at position \(x\):

The plot shows a circle in a vertical x y plane of radius 2.
In front there is half of an ellipse in the x z plane meeting the first circle
on the x axis. Inside there is vertical equilateral triangle whose base
is in the x y plane with its top and bottom vertices in the circle.

\(V=\dfrac{32}{3}\sqrt{3}\)

Here are the base with one cross-section and an accumulation of cross sections to form the final solid:

The video shows the circle and half ellipse of the previous plot.
It is being filled by equailateral triangular slices. — Accumulating Cross Sections
and the Solid

The circle is \(x^2+y^2=4\). So the endpoints of the base of the triangle are at \(y_\pm=\pm\sqrt{4-x^2}\). So its length is \(b=y_+-y_-=2\sqrt{4-x^2}\). The altitude is \(\sqrt{3}\) times half the base: \(h=\sqrt{3}\sqrt{4-x^2}\).
(Think about the \(30^\circ-60^\circ-90^\circ\) which is half of an equilateral triangle.)
So the area of the equilateral triangle is \[\begin{aligned} A&=\dfrac{1}{2}bh =\dfrac{1}{2}2\sqrt{4-x^2}\sqrt{3}\sqrt{4-x^2} \\ &=\sqrt{3}(4-x^2) \end{aligned}\]

The plot shows an equilateral triangle with one edge vetical at the
right. The sides have length 2. There is also a horizontal altitude
of length square root of 3.

Integrating this crosssectional area gives the volume: \[\begin{aligned} V&=\int_{-2}^2 A(x)\,dx =\int_{-2}^2 \sqrt{3}(4-x^2)\,dx \\ &=\sqrt{3}\left[ 4x-\dfrac{x^3}{3}\right]_{-2}^2 \\ &=\sqrt{3}\left[ 8-\dfrac{8}{3}\right]-\sqrt{3}\left[-8+\dfrac{8}{3}\right] =\dfrac{32}{3}\sqrt{3} \end{aligned}\]

PY: Checked Alt Text to here,

You can practice computing Volumes by Slicing by using the following Maplet (requires Maple on the computer where this is executed):

Volume by Slicing Rate It

13. Volume

a1. Volume by Slicing

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