find the volume of the solid whose base is the semicircle 𝑦

To find the volume of the solid whose base is the semicircle $y = 1 - x^2$ with $-1 \leq x \leq 1$, and whose cross-sections perpendicular to the $x$-axis are squares, we use the cross-section method.

Step 1: Identify the Side Length of the Square

Each cross-section is a square with a side length equal to the function $y = 1 - x^2$. This means the area of the square at any given $x$ is:

$A(x) = (\text{side length})^2 = (1 - x^2)^2$

Step 2: Set Up the Integral

To find the volume, we integrate the area function over the given interval:

$V = \int_{-1}^{1} A(x) \,dx = \int_{-1}^{1} (1 - x^2)^2 \,dx$

Step 3: Expand the Integrand

Expanding $(1 - x^2)^2$:

$(1 - x^2)^2 = 1 - 2x^2 + x^4$

Thus, the integral simplifies to:

$V = \int_{-1}^{1} (1 - 2x^2 + x^4) \,dx$

Step 4: Compute the Integral

We integrate term by term:

$\int_{-1}^{1} 1 \,dx = x \Big|_{-1}^{1} = 1 - (-1) = 2$

$\int_{-1}^{1} 2x^2 \,dx = 2 \int_{-1}^{1} x^2 \,dx$

Since $\int_{-a}^{a} x^n \,dx$ is zero for odd powers and symmetric for even powers:

$\int_{-1}^{1} x^2 \,dx = \frac{x^3}{3} \Big|_{-1}^{1} = \frac{1}{3} - \frac{(-1)^3}{3} = \frac{1}{3} - (-\frac{1}{3}) = \frac{2}{3}$

Thus,

$\int_{-1}^{1} 2x^2 \,dx = 2 \times \frac{2}{3} = \frac{4}{3}$

$\int_{-1}^{1} x^4 \,dx = \frac{x^5}{5} \Big|_{-1}^{1} = \frac{1}{5} - (-\frac{1}{5}) = \frac{2}{5}$

Step 5: Compute the Final Volume

$V = 2 - \frac{4}{3} + \frac{2}{5}$

Finding a common denominator (15):

$V = \frac{30}{15} - \frac{20}{15} + \frac{6}{15} = \frac{16}{15}$

Final Answer:

$V = \frac{16}{15}$

Let me know if you have any questions!

Related Problems:

1. Find the volume when the cross-sections are equilateral triangles.
2. Solve the problem when the cross-sections are semicircles.
3. Find the centroid of the semicircle base.
4. Determine the volume if the base was a full circle instead.
5. Find the volume using cylindrical shells (alternative approach).

Tip:

When dealing with square cross-sections, the function defining the region usually represents the side length, so remember to square it before integrating.