To calculate the volume of the solid formed when the area under $y = x^2$ between $x = 1$ and $x = 12$ is rotated about the x-axis, we use the formula for the volume of revolution:

$V = \pi \int_a^b [f(x)]^2 \, dx$

Here, $f(x) = x^2$, $a = 1$, and $b = 12$. Substituting $f(x)$ into the formula:

$V = \pi \int_1^{12} (x^2)^2 \, dx$

Simplify $(x^2)^2$ to $x^4$:

$V = \pi \int_1^{12} x^4 \, dx$

The integral of $x^4$ is:

$\int x^4 \, dx = \frac{x^5}{5}$

Apply the limits of integration:

$V = \pi \left[ \frac{x^5}{5} \right]_1^{12}$

Substitute $x = 12$ and $x = 1$:

$V = \pi \left( \frac{12^5}{5} - \frac{1^5}{5} \right)$

Calculate $12^5$ and $1^5$:

$12^5 = 248832, \quad 1^5 = 1$

Substitute these values:

$V = \pi \left( \frac{248832}{5} - \frac{1}{5} \right)$

Simplify:

$V = \pi \left( \frac{248831}{5} \right)$

$V = \frac{248831\pi}{5}$

Thus, the volume of the solid is:

$V = \frac{248831\pi}{5} \quad \text{(exact form)} \quad \text{or approximately } 156097.5 \, \pi \, \text{cubic units.}$

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Related Questions

1. What is the volume if the same region is rotated about the y-axis instead?
2. How does the volume change if $y = x^2$ is replaced with $y = x^3$?
3. Can this integral be solved numerically if a function doesn’t have a closed-form integral?
4. How would this problem change if bounds were different, say $x = 2$ to $x = 10$?
5. What is the surface area of the solid formed by rotating $y = x^2$ about the x-axis?

Tip: When rotating a curve about an axis, ensure you square the function for volume integrals and adjust limits carefully.