evaluate $\iiint\limits_{B}^{} \cosh \left ( x+y+z \right ) dxdydz$, where $B=\left \{ \left ( x,y,z \right ) \in R^3|x^2+y^2+z^2\le 1 \right \} $

Question

Evaluate $\iiint\limits_{B}^{} \cosh \left ( x+y+z \right ) dxdydz$, where $B=\left \{ \left ( x,y,z \right ) \in R^3|x^2+y^2+z^2\le 1 \right \} $

Since B is a unit ball which centers at the origin, I think it should be highly usefull to change coordinates to spherical coordinates. And of course, by definition, $$\cosh\left ( x+y+z\right ) = \frac{e^{x+y+z}+e^{-x-y-z}}{2}$$ But after changing coordinates, I still know little about how to integral such thing like......$$\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{1} \cosh \left ( r\sin \phi \cos \theta+r\sin \phi \sin \theta+r\cos \phi \right )r^2\sin \phi drd\phi d\theta $$ or$$\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{1} \frac{e^{\left ( r\sin \phi \cos \theta+r\sin \phi \sin \theta+r\cos \phi \right )}+e^{\left ( -r\sin \phi \cos \theta-r\sin \phi \sin \theta-r\cos \phi \right )}}{2} r^2\sin \phi drd\phi d\theta $$

Could anyone give some hints how I can do it please?

Answer 1

Consider this change of variables:

• $X=\frac1{\sqrt2}(x-y)$
• $Y=\frac1{\sqrt6}(x+y-2z)$
• $Z=\frac1{\sqrt3}(x+y+z)$

Note that $(x,y,z)\mapsto\left(\frac1{\sqrt2}(x-y),\frac1{\sqrt6}(x+y-2z),\frac1{\sqrt3}(x+y+z)\right)$ is a linear map and, more precisely, it is an orthogonal transformation; in particular, the absolute value of its determinant is equal to $1$, it is injective, and it maps $B$ onto $B$.

So, after doing this change of variable, your integral becomes$$\iiint_B\cosh\left(\sqrt3Z\right)\,\mathrm dX\,\mathrm dY\,\mathrm dZ.$$Now, using spherical coordinates, this becomes\begin{multline}\int_0^{2\pi}\int_0^\pi\int_0^1\cosh\left(\sqrt3\rho\cos\varphi\right)\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta=\\=2\pi\int_0^\pi\int_0^1\cosh\left(\sqrt3\rho\cos\varphi\right)\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi.\end{multline}But\begin{multline}\int_0^1\cosh\left(\sqrt3\rho\cos\varphi\right)\rho^2\sin\varphi\,\mathrm d\rho=\\=\frac1{18}\sec\varphi\bigl(\sqrt3(3\cos(2\varphi)+7)\sec\varphi\sinh\left(\sqrt{3}\cos\varphi\right)-12\cosh\left(\sqrt3\cos\varphi\right)\bigr)\tan\varphi\end{multline}and\begin{multline}\int_0^\pi\sec\varphi\bigl(\sqrt3(3\cos(2\varphi)+7)\sec\varphi\sinh\left(\sqrt{3}\cos\varphi\right)-12\cosh\left(\sqrt3\cos\varphi\right)\bigr)\tan\varphi\,\mathrm d\varphi=\\=12\cosh\left(\sqrt3\right)-4\sqrt3\sinh\left(\sqrt3\right).\end{multline}Therefore, your integral is equal to$$\frac\pi9\left(12\cosh\left(\sqrt3\right)-4\sqrt3\sinh\left(\sqrt3\right)\right).$$

Answer 2

Fig. 1 : The unit ball modelized as a pile of cylinders with common axis (D) with equation (2) (in blue) and infinitesimal height. A plane $P_k$ with equation (1) which is represented by its trace, a red triangle, on the coordinate planes ; its intersection with the unit ball is a disk. The coordinate axes are represented in black.

Let us consider planes $P_k$ with equation :

$$x+y+z=k\tag{1}$$

They share a common orthogonal line $(D)$ with equation $$x=y=z\tag{2}$$

• Plane $P_k$ intersects line $(D)$ in point $I_k:=(k/3,k/3,k/3)$

• Plane $P_k$ intersects the unit ball along a circle with center $I_k$ and radius $r(k)$ such that (Pythagorean theorem) :

$$(OI_k)^2+r(k)^2=1 \ \iff \ r(k)^2=1-k^2/3$$

Remark : one must have $-\sqrt{3} \le k \le \sqrt{3}$ in order that plane $P_k$ intersects the unit ball.

Therefore, if we consider the unit ball as represented by an "almost spherical" paper lantern made of a "pile" of cylinders with infinitesimal height $dk$ having base disks with radius $r(k)$, the triple integral is reduced to this single integral :

$$\int_{k=-\sqrt{3}}^{k=+\sqrt{3}}\cosh(k)\underbrace{(\pi r(k)^2)}_{\text{disk area}}\color{red}{\frac{1}{\sqrt{3}}}dk$$

$$=\frac{4\pi}{3 \sqrt{3}} \left(\sqrt{3}\cosh\left(\sqrt{3}\right)-\sinh\left(\sqrt{3}\right)\right) \approx 5.5878$$

finding back the result of José Carlos Santos (computation made by WA).

The presence of $\color{red}{\frac{1}{\sqrt{3}}}$ has to be explained : when we move from $k$ to $k+dk$, dist$(I_k,I_{k+dk})$ is the diagonal of a cube whose sides are $dk/\sqrt{3}$.

Many thanks to @Robert Z who has pointed a big double error of mine.

(for interested readers) Here is the Matlab which has generated the figure ; please note that the pile of cylinders has initialy been obtained with the $z$ axis as common axis, then, all them have undergone a rotation with angle $\theta=\operatorname{atan}(\sqrt{2})$ and axis $(-1,1,0)$ in order that the $z$ axis becomes the $(D)$ axis. This rotation is obtained as the exponential of a certain antisymmetrical matrix.

 clear all;close all;axis equal,axis off;hold on;
 L='linewidth';
 t=-1:0.01:1;
 k=9;T=round(k*t)/k;
 r=sqrt(1-T.^2);[X,Y,Z]=cylinder(r);
 a=1/sqrt(2);th=atan(sqrt(2));
 R=expm(th*[0,0,a;0,0,a;-a,-a,0]); % rotation matrix
 [p,q]=size(X); % p=201 ; q=21
 U=[X(:),Y(:),Z(:)]';S=R*U;
 X1=reshape(S(1,:),p,q);
 Y1=reshape(S(2,:),p,q);
 Z1=reshape(S(3,:),p,q);
 surf(X1,Y1,Z1);view(70,30);
 u=1.5;v=2.5;w=1.5;
 plot3(u*[0,1],u*[0,1],u*[0,1],L,5); % line x=y=z
 plot3([v,0,0,0,0],[0,0,v,0,0],[0,0,0,0,v],'k',L,2) % axes
 I=eye(3);T=w*I(:,[1,2,3,1])
 plot3(T(1,:),T(2,:),T(3,:),'r',L,5); % triangle

Answer 3

Computing the Integral via a Change of Coordinates

We have the orthonormal change of coordinates $$ \begin{align} u&=\frac{x+y+z}{\sqrt3}\\ v&=\frac{x-z}{\sqrt2}\\ w&=\frac{x-2y+z}{\sqrt6} \end{align}\tag1 $$ Then the integral becomes $$ \begin{align} \int_B\cosh(x+y+z)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z &=\int_B\cosh\left(u\sqrt3\right)\,\mathrm{d}v\,\mathrm{d}w\,\mathrm{d}u\tag{2a}\\ &=\int_{-1}^1\pi\left(1-u^2\right)\cosh\left(u\sqrt3\right)\,\mathrm{d}u\tag{2b}\\ &=\frac{2\pi}{\sqrt3}\int_{-1}^1u\sinh\left(u\sqrt3\right)\,\mathrm{d}u\tag{2c}\\ &=\frac{4\pi}3\cosh\left(\sqrt3\right)-\frac{2\pi}3\int_{-1}^1\cosh\left(u\sqrt3\right)\,\mathrm{d}u\tag{2d}\\ &=\frac{4\pi}3\cosh\left(\sqrt3\right)-\frac{4\pi}{3\sqrt3}\sinh\left(\sqrt3\right)\tag{2e}\\ &=\frac{4\pi}3\left(\cosh\left(\sqrt3\right)-\frac{\sinh\left(\sqrt3\right)}{\sqrt3}\right)\tag{2f} \end{align} $$ Explanation:
$\text{(2a):}$ apply $(1)$
$\text{(2b):}$ integrate in $v$ and $w$;
$\phantom{\text{(2b):}}$ since $u^2+v^2+w^2\le1$, for a given $u$,
$\phantom{\text{(2b):}}$ $v^2+w^2\le1-u^2$: a disk of radius $\sqrt{1-u^2}$
$\text{(2c):}$ integrate by parts
$\text{(2d):}$ integrate by parts
$\text{(2e):}$ integrate
$\text{(2f):}$ simplify

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Motivation for the Change of Coordinates

What we want to get is an orthonormal set of vectors. We start with an orthogonal set of vectors, and then normalize each one.

The first vector is $(1,1,1)$ because we want $x+y+z$ to be a variable; that is because $\cosh(x+y+z)$ is the integrand.

One choice for an orthogonal vector is something like $(1,0,-1)$; that is, $$ (1,1,1)\cdot(1,0,-1)=0\tag3 $$ For the third vector, because we have $(1,0,-1)$, we want the first and last entries equal; that is, a vector like $(1,\alpha,1)$, since $$ (1,0,-1)\cdot(1,\alpha,1)=0\tag4 $$ We choose $\alpha$ so that $$ 0=(1,1,1)\cdot(1,\alpha,1)=2+\alpha\tag5 $$ That is, we have $(1,-2,1)$.

Now we normalize, so that they all have norm $1$: $$ \begin{array}{c} \displaystyle\frac{(1,1,1)}{\sqrt3}\\ \displaystyle\frac{(1,0,-1)}{\sqrt2}\\ \displaystyle\frac{(1,-2,1)}{\sqrt6} \end{array}\tag6 $$ This leads to the change of variables in $(1)$.