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The sphere: $x^2+y^2+z^2\leq a^2$ is intercepted by the cylindrical surface $x^2+y^2=ax$. Calculate the intercepted volume.

Consider the intercepted volume of the upper hemisphere, and then multiply it by 2: $$D=\{(x,y):x^2+y^2\leq ax\}$$

Now calculate $A$: $$A=\iint\limits_{D}\sqrt{a^2-x^2-y^2}\text{d}x\mathrm{d}y=\int_{\color{magenta}{-\pi/2}}^{\pi/2}{\text{d}\theta\int^{a\cos\theta}_{0}}r\sqrt{a^2-r^2}\text{d}r \\= \color{magenta}{2}\int_{\color{magenta}{0}}^{\pi/2}\text{d}\theta\int^{a\cos\theta}_{0}r\sqrt{a^2-r^2}\text{d}r\\ =\frac{2}{3}a^3\int^{\pi/2}_{\color{magenta}{0}}(1-\sin^3\theta)\text{d}\theta=\color{red}{\frac{a^3}{3}(\pi-\frac{4}{3})} $$ if I don't change the lower limit of integral, the result would be: $$ \frac{1}{3}a^3\int^{\pi/2}_{\color{magenta}{-\pi/2}}(1-\sin^3\theta)\text{d}\theta= \color{red}{\frac{a^3}{3}\pi} $$

Why does the change in limit affect the result?(The two red parts are different)

Tim
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1 Answers1

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Thanks @Benjamin_Gal. In the comment of this question, he pointed out: $$ (\sin^2\theta)^{3/2}=|\sin^3\theta|\ne\sin^3\theta $$

And the correction would be: $$ \begin{eqnarray} A=\iint\limits_{D}\sqrt{a^2-x^2-y^2}\text{d}x\mathrm{d}y&=&\int_{\color{black}{-\pi/2}}^{\pi/2}{\text{d}\theta\int^{a\cos\theta}_{0}}r\sqrt{a^2-r^2}\text{d}r \\ &=&\frac{1}{3}a^3\int^{\pi/2}_{\color{black}{-\pi/2}}(1-|\sin^3\theta|)\text{d}\theta=\color{black}{\frac{a^3}{3}(\pi-\frac{4}{3})} \end{eqnarray} $$

Tim
  • 156