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What is the best method for evaluating the following double integral?

$$ \int_{0}^{b}\int_{0}^{x}\,\sqrt{\,a^{2} - x^{2} - y^{2}\,}\,\,{\rm d}y\,{\rm d}x\,, \qquad a > \sqrt{\,2\,}\,\,b $$

Is there exist an easy method?

My try:

$$\int_0^b\int_0^x\sqrt{a^2-x^2-y^2}dy\,dx=\int_0^{\frac{\pi}{4}}\int_0^{b\sec(\theta)}r\sqrt{a^2-r^2}dr\,d\theta$$ $$=\int_0^{\frac{\pi}{4}}\frac{-1}{3}\left[(a^2-r^2)\sqrt{a^2-r^2}\right]_0^{b\sec(\theta)}d\theta$$ $$=\frac{1}{3}\int_0^{\frac{\pi}{4}}\left[a^3-(a^2-b^2\sec^2(\theta))\sqrt{a^2-b^2\sec^2(\theta)}\right]d\theta$$ but evaluating above integral is very difficult and antiderivative is very complexity! see here.

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  • $\begingroup$ Polar coordinates looks promising. Do you know how to perform change of variables? $\endgroup$ – Mark Fantini Nov 20 '13 at 13:57
  • $\begingroup$ How did you arrive to those ugly limits? Were they given to you? $\endgroup$ – DonAntonio Nov 20 '13 at 13:58
  • $\begingroup$ @user95733: the point is that you want to integragte a function that behaves nicely on the circle, over a triangle. That's why the suspicion that maybe they are not the right limits. $\endgroup$ – Martin Argerami Nov 20 '13 at 15:12
  • $\begingroup$ It rather looks like integrating over some piece of a sphere : $x^2 + y^2 + z^2 = a^2$. $\endgroup$ – Han de Bruijn Nov 20 '13 at 15:14
  • $\begingroup$ @Han: you are right. A triangular section of a sphere. $\endgroup$ – Martin Argerami Nov 20 '13 at 15:22
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Suppose $$I=\int_0^b\int_0^x\sqrt{a^2-x^2-y^2}dy\,dx$$ Let $y=\sqrt{a^2-x^2}\sin \theta$ then $dy=\sqrt{a^2-x^2}\cos \theta\,d\theta$, so $$I=\int_0^b\int_0^{\arcsin\frac{x}{\sqrt{a^2-x^2}}}(a^2-x^2)\cos^2 \theta\,d\theta\,dx$$ $$=\frac{1}{2}\int_0^b(a^2-x^2)\left[\theta+\frac{1}{2}\sin 2\theta\right]_0^{\arcsin\frac{x}{\sqrt{a^2-x^2}}}\,dx$$ and now note that $\sin \theta=\frac{x}{\sqrt{a^2-x^2}}$, so $\cos \theta=\sqrt{\frac{a^2-2x^2}{a^2-x^2}}$ and $\frac{1}{2}\sin 2\theta=\sin \theta\cos \theta=\frac{x\sqrt{a^2-2x^2}}{a^2-x^2}$. therefore $$I=\frac{1}{2}\left(\int_0^b(a^2-x^2)\arcsin\frac{x}{\sqrt{a^2-x^2}}\,dx+\int_0^bx\sqrt{a^2-2x^2}\,dx\right)=\frac{1}{2}(I_1+I_2).$$ for evaluating $I_1$ use integrating by parts. If you let $u=\arcsin\frac{x}{\sqrt{a^2-x^2}}$ and $dv=(a^2-x^2)dx$, then $$du=\frac{a^2}{(a^2-x^2)\sqrt{a^2-2x^2}}dx,v=a^2x-\frac{x^3}{3}$$ therefore $$I_1=\frac{3a^2b-b^3}{3}\arcsin\frac{b}{\sqrt{a^2-b^2}}+I_3$$ and $$I_3=-\int_0^b\frac{a^2(a^2-\frac{x^2}{3})x}{(a^2-x^2)\sqrt{a^2-2x^2}}\,dx$$ now let $u=\sqrt{a^2-2x^2}$, then $du=\frac{-2x}{\sqrt{a^2-2x^2}}dx$ and $x^2=\frac{a^2-u^2}{2}$. so $$I_3=\frac{a^2}{2}\int_a^{\sqrt{a^2-2b^2}}\frac{a^2-\frac{a^2-u^2}{6}}{a^2-\frac{a^2-u^2}{2}}\,du=\frac{a^2}{6}\int_a^{\sqrt{a^2-2b^2}}\frac{u^2+5a^2}{u^2+a^2}\,du$$ $$=\frac{a^2}{6}\int_a^{\sqrt{a^2-2b^2}}\left(1+\frac{a^2}{u^2+a^2}\right)\,du=\frac{a^2}{6}(\sqrt{a^2-2b^2}-a)+\frac{2a^3}{3}\left(\arctan\frac{\sqrt{a^2-2b^2}}{a}-\frac{\pi}{4}\right)$$ for $I_2$ we have $$I_2=\int_0^bx\sqrt{a^2-2x^2}\,dx=\frac{-1}{6}\left[(a^2-2x^2)\sqrt{a^2-2x^2}\right]_0^b=\frac{1}{6}(a^3-(a^2-2b^2)\sqrt{a^2-2b^2})$$ Hence,

$$I=\frac{3a^2b-b^3}{6}\arcsin\frac{b}{\sqrt{a^2-b^2}}+\frac{a^3}{3}\arctan\frac{\sqrt{a^2-2b^2}-a}{\sqrt{a^2-2b^2}+a}+\frac{b^2}{6}\sqrt{a^2-2b^2}.$$

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Of course this is an exercise in integration. But nevertheless it can be done using elementary geometry alone. We are told to compute the volume of a body $K$ which is bounded by several planar faces and a peace of a spherical surface. The following figure shows the situation as seen from the tip of the $x$-axis. I have put $a=1$ and written $p$ instead of $b$.

enter image description here

The body $K$ contains (a) the pyramid $P$ with base the triangle with vertices $(0,0,0)$,$(p,0,0)$, $(p,p,0)$, and of height $\sqrt{1-2p^2}$. The volume of this pyramid is $${\rm vol}(P)={1\over6}p^2\sqrt{1-2p^2}\ .$$ Furthermore $K$ contains (b) part of a sector $S$ of central angle $$\alpha:=\arctan{p\over\sqrt{1-2p^2}}$$ of a spherical segment. The two radii of this segment are $1$ and $\sqrt{1-p^2}$, and its thickness is $p$. The volume of the full sector $S$ is therefore given by $${\rm vol}(S)={\alpha\over 2\pi}{\pi\over6} p\bigl(3+3(1-p^2) +p^2\bigr)={\alpha\over6}p(3-p^2)\ .$$ From the volume of $S$ we (c) have to deduct the volume of a triangular spherical sector $T$, whereby the angles of the spherical triangle in question (shaded in the figure) are ${\pi\over2}$, ${\pi\over4}$ and a certain $\beta$. One leg of this triangle is $\alpha$, and a standard formula for right spherical triangles then tells us that $$\cos\beta=\sin{\pi\over4}\cos\alpha={1\over\sqrt{2}}\cos\alpha\ .$$ The spherical area of the triangle is then $\beta-{\pi\over 4}$, so that $${\rm vol}(T)={1\over3}\bigl(\beta-{\pi\over 4}\bigr)\ .$$ Finally $${\rm vol}(K)={\rm vol}(P)+{\rm vol}(S)-{\rm vol}(T)\ ,$$ which maybe can be simplified somewhat.

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$\int_0^b\int_0^x\sqrt{a^2-x^2-y^2}~dy~dx$

$=\int_0^b\left[\dfrac{y\sqrt{a^2-x^2-y^2}}{2}+\dfrac{a^2-x^2}{2}\sin^{-1}\dfrac{y}{\sqrt{a^2-x^2}}\right]_0^x~dx$ (according to http://en.wikipedia.org/wiki/List_of_integrals_of_irrational_functions)

$=\int_0^b\dfrac{x\sqrt{a^2-2x^2}}{2}dx+\int_0^b\dfrac{a^2-x^2}{2}\sin^{-1}\dfrac{x}{\sqrt{a^2-x^2}}dx$

Can you take it from here?

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(The following is basically obsolete because already answered in https://math.stackexchange.com/questions/571967/integral-inta2-b2-sec2x-sqrta2-b2-sec2xdx)

The first integral of Harry Peter is easy with the substitution $x^2=t$. The second, attacked with partial integration, requires to integrate $(a^2-x^2)/2$ which is $x(3a^2-x^2)/6$ and to differentiate the $\arcsin$ which gives $a^2/\sqrt{a^2-2x^2}/(a^2-x^2)$. So we need besides the preintegrated term essentially $$\int dx \frac{x(3a^2-x^2)}{6}\frac{a^2}{\sqrt{a^2-2x^2}(a^2-x^2)}$$ which turns with $x^2=t$ in something proportional to $$\int dx \frac{(3a^2-t)}{6}\frac{a^2}{\sqrt{a^2-2t}(a^2-t)}$$. By expansion of $(3a^2-6)/(a^2-t)$ into partial fractions only the format $1/[\sqrt{a^2-2t}(a^2-t)]$ is required which seems with $y=a^2-2t$ to be covered by $$\int \frac{dy}{(a+by)\surd y} =\frac{2}{\sqrt{ab}}\arctan\sqrt{\frac{by}{a}}$$

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In a sense "Polar coordinates does not help much since the integrand likes them while the domain does not". It helps to know about partial integration, that is $$\int f'(x)g(x)dx = f(t)g(t)- \int f(x)g'(x)dx +C $$ In addition one should know about the derivatives of $\arcsin$ and $\arctan$.

Now to the integral, first we simplify \begin{eqnarray}\int_0^x\sqrt{a^2-x^2-y^2}dy &=& \int_0^x\sqrt{a^2-x^2}\sqrt{1-\left(\frac{y}{\sqrt{a^2-x^2}}\right)^2}dy \\ &=&\int_0^{x/\sqrt{a^2-x^2}}(a^2-x^2)\sqrt{1-t^2}dy\tag{1}\end{eqnarray} Next we use partial integration \begin{eqnarray}I=\int_0^\eta\sqrt{1-t^2}dt &=& \left[t\sqrt{1-t^2}\right]_0^\eta-\int_0^\eta t\cdot\left(\frac{-t}{\sqrt{1-t^2}}\right)dt\\&=& \eta\sqrt{1-\eta^2}+\int_0^\eta \frac{t^2}{\sqrt{1-t^2}}dt \end{eqnarray} and then $$\int_0^\eta \frac{t^2}{\sqrt{1-t^2}}dt = \int_0^\eta \frac{-(1-t^2)}{\sqrt{1-t^2}}dt+\int_0^\eta \frac{1}{\sqrt{1-t^2}}dt=-I +\arcsin{\eta}$$ Thus $$I=\frac12\left(\eta\sqrt{1-\eta^2} +\arcsin{\eta} \right)\tag{2}$$ which we use in (1) to get \begin{eqnarray} \int_0^{x/\sqrt{a^2-x^2}}(a^2-x^2)\sqrt{1-t^2}dy &=& \frac{(a^2-x^2)}2\Big(\frac{x}{\sqrt{a^2-x^2}}\sqrt{1-\frac{x^2}{a^2-x^2}} \\ &&+\arcsin{\frac{x}{\sqrt{a^2-x^2}}} \Big)\\ &=&\frac{x}{2}\sqrt{a^2-2x^2} +\frac{(a^2-x^2)}2\arcsin{\frac{x}{\sqrt{a^2-x^2}}} \end{eqnarray} It follows that $$\int_0^b \int_0^x\sqrt{a^2-x^2-y^2}dy\,dx= \int_0^b\frac{x}{2}\sqrt{a^2-2x^2}dx+ \int_0^b \frac{(a^2-x^2)}2\arcsin{\frac{x}{\sqrt{a^2-x^2}}}dx$$ and since $$\int_0^b\frac{x}{2}\sqrt{a^2-2x^2}dx= \frac{1}{12}(a^2-2b^2)^{3/2}$$ it suffice to calculate \begin{eqnarray}\int_0^b \frac{(a^2-x^2)}2\arcsin{\frac{x}{\sqrt{a^2-x^2}}}dx&=&\frac1{2a^2}\int_0^b \frac{(1-(x/a)^2)}2\arcsin{\frac{(x/a)}{\sqrt{1-(x/a)^2}}}dx\\ &=&\frac1{2a}\int_0^{b/a} (1-t^2)\arcsin{\frac{t}{\sqrt{1-t^2}}}dt\end{eqnarray} which can be calculated in the same fashion as we did with $I$ above - I leave this fun part to you.

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