Discussion
Turning the curve \( y = \sqrt{x} \) about the \( x \)-axis between \( x = 0 \) and \( x = 4 \) sweeps a solid measuring
Answer: D
With \( y^{2} = x \) the disc formula becomes an integral of \( x \) itself:
\[ V \;=\; \pi\int_{0}^{4} x\,dx \;=\; \pi\left[\frac{x^{2}}{2}\right]_{0}^{4} \;=\; 8\pi. \]
Forgetting to square the radius would leave \( \pi\int\sqrt{x}\,dx = 16\pi/3 \) instead, so the squaring is what makes the integrand elementary here
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