How to find volume with square cross sections from a region under curves

Expert intro: For square cross sections, the side length is just the height of the region at each x-value. Once that is clear, the volume is an area integral of the side squared.

Detailed walkthrough

Step 1: Find the side length of each square

Cross sections are perpendicular to the x-axis, so the side length equals the vertical distance from the x-axis to the top of region $S$.

That gives a piecewise side length:

\begin{cases} \frac12 x^2, & 0\le x\le 2 \\ 8-2x, & 2\le x\le 4 \end{cases}$$ ### Step 2: Square the side length Because each cross section is a square, the area is $s(x)^2$. $$A(x)= \begin{cases} \left(\frac12 x^2\right)^2=\frac14 x^4, & 0\le x\le 2 \\ (8-2x)^2, & 2\le x\le 4 \end{cases}$$ ### Step 3: Integrate over both intervals $$V=\int_0^2 \frac14 x^4\,dx+\int_2^4 (8-2x)^2\,dx$$ First part: $$\int_0^2 \frac14 x^4\,dx=\left[\frac{x^5}{20}\right]_0^2=\frac{32}{20}=\frac85$$ Second part: $$\int_2^4 (8-2x)^2\,dx=\int_2^4 (64-32x+4x^2)\,dx=\frac{32}{3}$$ ### Step 4: Add the results $$V=\frac85+\frac{32}{3}=\frac{184}{15}$$ ### Answer $$\boxed{\frac{184}{15}}$$ ### 💡 Pitfall guide The most common error is forgetting to square the side length. For square cross sections, the cross-sectional area is not just the height; it is height squared. ### 🔄 Real-world variant If the cross sections were semicircles instead of squares, the area formula would become $\frac{\pi}{8}s(x)^2$ instead of $s(x)^2$. Same side length, different cross-sectional shape, different volume. ### 🔍 Related terms cross-sectional area, piecewise function, definite integral