Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Polar coordinates looks promising. Do you know how to perform change of variables? –  Mark Fantini Nov 20 '13 at 13:57
    
How did you arrive to those ugly limits? Were they given to you? –  DonAntonio Nov 20 '13 at 13:58
    
@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. –  Martin Argerami Nov 20 '13 at 15:12
    
It rather looks like integrating over some piece of a sphere : $x^2 + y^2 + z^2 = a^2$. –  Han de Bruijn Nov 20 '13 at 15:14
    
@Han: you are right. A triangular section of a sphere. –  Martin Argerami Nov 20 '13 at 15:22

5 Answers 5

up vote 7 down vote accepted

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,

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

share|improve this answer

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.

share|improve this answer
    
You learn me a geometric method. Thanks very much. –  user95733 Nov 26 '13 at 11:56

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.

share|improve this answer

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

$=\int_0^b\left[\dfrac{y\sqrt{a^2-x^2-y^2}}{2}+\dfrac{a^2-x^2}{2}\sin^{-1}\dfrac{y}{\sqrt{a^2-x^2}}\right]_0^x~dx$ (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?

share|improve this answer

(The following is basically obsolete because already answered in Integral $\int(a^2-b^2\sec^2(x))\sqrt{a^2-b^2\sec^2(x)}dx$)

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 $$\int \frac{dy}{(a+by)\surd y} =\frac{2}{\sqrt{ab}}\arctan\sqrt{\frac{by}{a}}$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.