Set up the volume integral
Slabs that add up to a solid
A solid has squares for cross sections, perpendicular to the x-axis. Its base is the region between the x-axis and the line \(y = 3 - x\), from \(x = 0\) to \(x = 3\).
The faint outline is the whole solid, cut into 3 slabs of thickness 1. A teal block replaces each slab. Its face is the cross section at the middle of the slab, and its volume is that area times the thickness.
The faces at \(x = 0.5\), \(1.5\) and \(2.5\) have areas 6.25, 2.25 and 0.25, so the three blocks have a total volume of 8.75.
Press the buttons for more slabs. With 6 slabs, the total is 8.9375. Thinner slabs fit the solid more closely, so the total approaches 9.
Add the slabs
The slabs run along the x-axis, from a where the base starts to b where it ends.
Each face is a cross section, so the volume of a slab is close to the area of its face times its thickness.
Drag the slider to make the slabs thinner. The volume of the solid stays the same, and the total of the slabs approaches it.
Notation
The limits \(a\) and \(b\) are the smallest and largest \(x\) in the base, where the boundary curves meet or where the region ends.
The slab at \(x\) has a face of area \(\cb{A(x)}\), the area of the cross section there, not of the base. Its thickness is \(\Delta x\), so its volume is close to their product.
\[\Delta V \approx \cb{A(x)}\,\Delta x\]You build \(A(x)\) from the base width, as in the last skill.
Add the volumes of the slabs from \(a\) to \(b\). As \(\Delta x\) approaches 0, the sum becomes a definite integral.
\[V = \int_a^b \cb{A(x)}\,dx\]If the cross sections are perpendicular to the y-axis, the slabs stack along the y-axis. Use \(A(y)\) and \(dy\), with y-values as limits.
When it fails
The limits must cover the whole base. Press the button to stop the stack of slabs before the right end of the base.
The faint part of the solid has no slabs, so the sum is too small. The formula for \(A(x)\) gives a number outside the base too. Read the limits from the base.
Example 1: squares on a root curve
The base of a solid is the region between the x-axis and \(y = \sqrt{x + 1}\), for \(0 \le x \le 3\). Its cross sections perpendicular to the x-axis are squares. Write an integral for the volume.
The lower boundary is the x-axis, so the side of each square is the height of the curve.
\[\cy{w(x)} = \sqrt{x + 1}\]The area of a square is its side squared, and squaring a square root removes the root. As a check, at \(x = 3\) the side is 2 and the area is 4, which equals \(3 + 1\).
\[\cb{A(x)} = \left(\sqrt{x + 1}\right)^2 = x + 1\]The base is given for \(0 \le x \le 3\), so these are the limits.
\[V = \int_0^3 (x + 1)\,dx\]
Example 2: slices perpendicular to the y-axis
The base of a solid is the triangle bounded by the two axes and the line \(x + 2y = 8\). Its cross sections perpendicular to the y-axis are squares. Write an integral for the volume.
The slices are perpendicular to the y-axis, so each base runs from the y-axis, at \(x = 0\), to the line. Solve the line for \(x\).
\[\cy{w(y)} = 8 - 2y\]The cross sections are squares, so square the side.
\[\cb{A(y)} = (8 - 2y)^2\]A tempting wrong move: The line meets the x-axis at 8, so use 0 and 8 as limits.
\[\cr{V = \int_0^8 (8 - 2y)^2\,dy}\]The red 8 is an x-value, but the variable is \(y\). The base reaches only to \(y = 4\), so limits up to 8 add slabs beyond it, where there is no base.
The slabs stack along the y-axis, so the limits are y-values. At \(x = 0\) the line is at \(y = 4\), so the base runs from \(y = 0\) to \(y = 4\).
\[V = \int_0^4 (8 - 2y)^2\,dy\]
Practice
The base of a solid is the region between \(y = \frac{1}{x}\) and \(y = \frac{5}{2} - x\). Its cross sections perpendicular to the x-axis are rectangles of height 2. Write the volume as an integral. Do not evaluate it.
The base begins and ends where the two curves meet.
Solution
The curves meet where \(\frac{1}{x} = \frac{5}{2} - x\). Multiply both sides by \(x\). The quadratic factors, and its roots are the ends of the base, so they are the limits.
\[\begin{gathered} x^2 - \tfrac{5}{2}x + 1 = 0, \\ \left(x - \tfrac{1}{2}\right)(x - 2) = 0, \\ x = \tfrac{1}{2} \text{ and } x = 2 \end{gathered}\]At \(x = 1\), the line is at height \(\frac{3}{2}\) and the curve is at height 1, so the line is the upper curve. Subtract the lower curve.
\[\cy{w(x)} = \tfrac{5}{2} - x - \tfrac{1}{x}\]The face is a rectangle, so its area is the base times the height 2. Integrate between the limits.
\[V = \int_{1/2}^{2} 2\left(\tfrac{5}{2} - x - \tfrac{1}{x}\right)dx\]
Questions
The base of a solid is the region between the x-axis and \(y = x^3\), for \(0 \le x \le 2\). Its cross sections perpendicular to the x-axis are squares.
The slabs stop at x = 1, and the faint part of the solid has none. What upper limit covers the whole solid?
- 2
- 1
- 8
Answer
2
Which integral gives the volume of the solid?
- \(\displaystyle\int_0^2 x^6\,dx\)
- \(\displaystyle\int_0^2 x^3\,dx\)
- \(\displaystyle\int_0^8 x^6\,dx\)
Answer
\(\displaystyle\int_0^2 x^6\,dx\)
The base lies between the y-axis and \(x = \ln y\), for \(1 \le y \le e\). The cross sections are squares perpendicular to the y-axis. Which integral gives the volume?
- \(\displaystyle\int_1^e (\ln y)^2\,dy\)
- \(\displaystyle\int_0^1 (\ln y)^2\,dy\)
- \(\displaystyle\int_1^e \ln y\,dy\)
Answer
\(\displaystyle\int_1^e (\ln y)^2\,dy\)
Summary
You write the volume as the integral of the cross-section area, with the limits at the two ends of the base.
This works because each thin slab has volume close to \(A(x)\,\Delta x\), and the integral adds the slabs.
You use it when you know the area of every cross section.
This is one of four skills in topic 8.7 of Which & Why: AP Calculus.