Let S S S be a solid and suppose that the area of the cross section in the plane perpendicular to the
x − x- x − axis is A ( x ) A(x) A ( x ) for a ≤ x ≤ b . a\leq x\leq b. a ≤ x ≤ b .
V = ∫ a b A ( x ) d x V=\displaystyle\int_{a}^bA(x)dx V = ∫ a b A ( x ) d x
V = ∫ 0 1 / 2 2 1 − 4 x 2 d x = ∫ 0 1 / 2 1 1 4 − x 2 d x 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 V = ∫ 0 1/2 1 − 4 x 2 2 d x = ∫ 0 1/2 4 1 − x 2 1 d x Trigonometric substitution
x = 1 2 sin t , − π 2 ≤ t ≤ π 2 x=\dfrac{1}{2}\sin t, -\dfrac{\pi}{2}\leq t\leq \dfrac{\pi}{2} x = 2 1 sin t , − 2 π ≤ t ≤ 2 π
d x = 1 2 cos t d t dx=\dfrac{1}{2}\cos t\ dt d x = 2 1 cos t d t
1 4 − x 2 = 1 2 cos t \sqrt{{1 \over 4}-x^2}=\dfrac{1}{2}\cos t 4 1 − x 2 = 2 1 cos t
∫ 1 1 4 − x 2 d x = ∫ 1 2 cos t 1 2 cos t d t = t + C = arcsin ( 2 x ) + C \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 ∫ 4 1 − x 2 1 d x = ∫ 2 1 cos t 2 1 cos t d t = t + C = arcsin ( 2 x ) + C
V = ∫ 0 1 / 2 2 1 − 4 x 2 d x = ∫ 0 1 / 2 1 1 4 − x 2 d x = 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= V = ∫ 0 1/2 1 − 4 x 2 2 d x = ∫ 0 1/2 4 1 − x 2 1 d x =
= [ arcsin ( 2 x ) ] 1 / 2 0 = π 2 − 0 = π 2 ( u n i t s 3 ) =[\arcsin(2x)]\begin{matrix}
1/2 \\
0
\end{matrix}={\pi \over 2}-0={\pi \over 2} (units^3) = [ arcsin ( 2 x )] 1/2 0 = 2 π − 0 = 2 π ( u ni t s 3 )
V = π 2 c u b i c u n i t s V={\pi \over 2}\ cubic\ units V = 2 π c u bi c u ni t s
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