Question
Cross-sections of the solid $S$ in planes perpendicular to the $x$ -axis are circles with diameters extending from the curve $y=\frac{1}{2} \sqrt{x}$ to the curve $y=\sqrt{x}$ for $0 \leqslant x \leqslant 4$.
Step 1
First, we need to find the radius of the circle in terms of x. Since the diameter of the circle is the distance between the two curves, we can find the radius by taking half of the difference between the y-values of the two curves. Show more…
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Key Concepts
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The solid $S$ is bounded by circles that are perpendicular to the $x$ -axis, intersect the $x$ -axis, and have centers on the parabola $y=\frac{1}{2}\left(1-x^{2}\right),-1 \leqslant x \leqslant 1$.
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The solid lies between planes perpendicular to the $x$ -axis at $x=0$ and $x=4 .$ The cross-sections perpendicular to the axis on the interval $0 \leq x \leq 4$ are squares whose diagonals run from the parabola $y=-\sqrt{x}$ to the parabola $y=\sqrt{x}$ .
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