Answer to Question #117677 in Calculus for Lizwi

Question

Find the volume of the solid with cross-sectional area 2/√(1−4x^2) lying between x=0 and x=12

Answer 1

Let $S$ be a solid and suppose that the area of the cross section in the plane perpendicular to the

 $x-$axis is $A(x)$ for $a\leq x\leq b.$

$V=\displaystyle\int_{a}^bA(x)dx$

$V=\displaystyle\int_{0}^{1/2}{2\over \sqrt{1-4x^2}}dx=\displaystyle\int_{0}^{1/2}{1\over \sqrt{{1 \over 4}-x^2}}dx$

Trigonometric substitution

$x=\dfrac{1}{2}\sin t, -\dfrac{\pi}{2}\leq t\leq \dfrac{\pi}{2}$

$dx=\dfrac{1}{2}\cos t\ dt$

$\sqrt{{1 \over 4}-x^2}=\dfrac{1}{2}\cos t$

$\int{1\over \sqrt{{1 \over 4}-x^2}}dx=\int{\dfrac{1}{2}\cos t\over \dfrac{1}{2}\cos t}dt=t+C=\arcsin(2x)+C$

$V=\displaystyle\int_{0}^{1/2}{2\over \sqrt{1-4x^2}}dx=\displaystyle\int_{0}^{1/2}{1\over \sqrt{{1 \over 4}-x^2}}dx=$

$=[\arcsin(2x)]\begin{matrix} 1/2 \\ 0 \end{matrix}={\pi \over 2}-0={\pi \over 2} (units^3)$

$V={\pi \over 2}\ cubic\ units$