Volumes with Cross Sections: Squares and Rectangles
Introduced| English |
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| cross sections/krɒs ˈsekʃnz/ |
Stacking up thin slabs into a solid
- Integration also finds volumes, not just areas — by adding up thin slices of a solid.
- Slice a solid perpendicular to an axis; each slice is a thin slab of some cross-sectional area $A(x)$.
- Add them up: $V=\displaystyle\int_a^b A(x)\,dx$ — the integral of the cross-sectional area.
- This lesson covers solids whose cross sections 横截面 are squares or rectangles.
The general volume formula
- If the cross section at $x$ has area $A(x)$ and thickness $dx$, its tiny volume is $A(x)\,dx$.
- Sum over the solid: $$V=\int_a^b A(x)\,dx$$
- Everything reduces to: write the cross-sectional area as a function of $x$, then integrate.
- The hard part is the geometry of one slice, not the calculus.
The volume of a solid with cross-sectional area $A(x)$ is...
Add up slabs $A(x)\,dx$.
To use $V=\int_a^b A(x)\,dx$, the cross-section slices must be...
Slices are perpendicular to the axis you integrate along.
Square cross sections
- A common setup: cross sections are squares with base sitting on the region between two curves.
- If the base length is $s(x)$ (often $f(x)-g(x)$, the gap between curves), a square's area is $s(x)^2$.
- So $V=\displaystyle\int_a^b \big(s(x)\big)^2\,dx$.
- The base of each square spans the region; its area is base-squared.
The base region under √x
y = a·√x
Each square slice sits on the base; its side is the height $\sqrt{x}$, so its area is $(\sqrt{x})^2=x$.
For square cross sections with base $s(x)$, the area is...
A square of side $s$ has area $s^2$.
The base length of a cross section is often the distance between the boundary curves, $f(x)-g(x)$.
The base spans the region between the curves.
Rectangular cross sections
- For rectangles, the area is base $\times$ height: $A(x)=s(x)\cdot h(x)$.
- The height might be a fixed multiple of the base, or given separately in the problem.
- $V=\displaystyle\int_a^b s(x)\,h(x)\,dx$.
- Read the problem carefully to get the height rule right.
Base under $y=\sqrt x$ on $[0,4]$, square cross sections: $A(x)=x$. Find $V=\int_0^4 x\,dx$.
$\big[\tfrac{x^2}{2}\big]_0^4=8$.
For rectangular cross sections, area = base ____ height.
$A=s\cdot h$ for a rectangle.
The base length $s(x)$ is usually the distance between the boundary curves ($f(x)-g(x)$) or between a curve and an axis — figure out what spans the base before squaring. For squares, the area is $s(x)^2$ (don't forget to square the whole gap). Slices must be perpendicular to the axis you integrate along.
A solid has base the region under $y=\sqrt{x}$ from $x=0$ to $4$, with square cross sections perpendicular to the $x$-axis.
- The base of each square is $s(x)=\sqrt{x}$ (curve to $x$-axis).
- Cross-sectional area: $A(x)=(\sqrt{x})^2=x$.
- $V=\displaystyle\int_0^4 x\,dx=\Big[\tfrac{x^2}{2}\Big]_0^4=8$.
A solid's volume is $V=\int_a^b A(x)\,dx$, the integral of its cross-sectional area. For square cross sections with base $s(x)$, $A(x)=s(x)^2$; for rectangles, $A(x)=s(x)\cdot h(x)$. Find the base $s(x)$ (often the gap between curves), write $A(x)$, and integrate.