Math Genius: How to solve $int_0^2 int_0^sqrt{4-x^{2}} int_0^sqrt{4-x^2 -y^2} z sqrt{4-x^2 -y^2} , dz , dy , dx$ in spherical coordinate

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$$int_0^2 int_0^sqrt{4-x^{2}} int_0^sqrt{4-x^2 -y^2} z sqrt{4-x^2 -y^2} , dz , dy , dx$$

The task is to solve this integral using spherical coordinate. After I tried to change the variable, I got
$$ int _0^{frac{pi }{2}}int _0^{frac{pi }{2}}int _0^2left(rho :cosleft(phi right)sqrt{4-rho ^2left(sinleft(phi right)right)^2}right):rho ^2sinleft(phi right)drho :dtheta :dphi
$$

Which I think pretty ugly with $sqrt{4-rho ^2left(sinleft(phi right)right)^2}$ . Is there anything I did wrong on the variable changing process? If it’s not, what are the approaches to solve this integral?

The integral simplifies like so

$$int_0^{2}int_0^{frac{pi}{2}}int_0^{frac{pi}{2}}rho^3sinphicosphisqrt{4-rho^2sin^2phi}:dtheta:dphi:drho = frac{pi}{4}int_0^{2}int_0^{frac{pi}{2}} rhosqrt{4-rho^2sin^2phi} :d(rho^2sin^2phi):drho$$

$$= frac{pi}{6}int_0^2 -rho left[4-rho^2sin^2phiright]^{frac{3}{2}}biggr|_0^{frac{pi}{2}}:drho = frac{pi}{6}int_0^2 8rho-rho(4-rho^2)^{frac{3}{2}}:drho = frac{8pi}{5}$$

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Math Genius: How to solve $int_0^2 int_0^sqrt{4-x^{2}} int_0^sqrt{4-x^2 -y^2} z sqrt{4-x^2 -y^2} , dz , dy , dx$ in spherical coordinate

Original Source Link

$$int_0^2 int_0^sqrt{4-x^{2}} int_0^sqrt{4-x^2 -y^2} z sqrt{4-x^2 -y^2} , dz , dy , dx$$

The task is to solve this integral using spherical coordinate. After I tried to change the variable, I got
$$ int _0^{frac{pi }{2}}int _0^{frac{pi }{2}}int _0^2left(rho :cosleft(phi right)sqrt{4-rho ^2left(sinleft(phi right)right)^2}right):rho ^2sinleft(phi right)drho :dtheta :dphi
$$

Which I think pretty ugly with $sqrt{4-rho ^2left(sinleft(phi right)right)^2}$ . Is there anything I did wrong on the variable changing process? If it’s not, what are the approaches to solve this integral?

The integral simplifies like so

$$int_0^{2}int_0^{frac{pi}{2}}int_0^{frac{pi}{2}}rho^3sinphicosphisqrt{4-rho^2sin^2phi}:dtheta:dphi:drho = frac{pi}{4}int_0^{2}int_0^{frac{pi}{2}} rhosqrt{4-rho^2sin^2phi} :d(rho^2sin^2phi):drho$$

$$= frac{pi}{6}int_0^2 -rho left[4-rho^2sin^2phiright]^{frac{3}{2}}biggr|_0^{frac{pi}{2}}:drho = frac{pi}{6}int_0^2 8rho-rho(4-rho^2)^{frac{3}{2}}:drho = frac{8pi}{5}$$

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Math Genius: Finding $iiint_K (x^2 -z^2), dx, dy ,dz$

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Calculate the integral
$$
iiint_K (x^2 -z^2), dx, dy ,dz,,
$$

where $$K=left{(x,y,z):x,y,zge 0,x+y+zle 1right}$$

I have tried solving it but I really don’t understand how to handle the upper limits. I really need some guidance.

Update:
I got it to zero by using it’s symmetry as given in comments. Thanks for the help. I also confirmed it by calculating it myself with the values inputted.

The region has an $xz$ exchange symmetry, i.e.

$$(x,y,z) in K implies (z,y,x) in K$$

In other words

$$I = iiint_K f(x,y,z):dV = iiint_K f(z,y,x):dV$$

but in this problem for $f(x,y,z) = x^2-z^2$ we have that

$$f(z,y,x) = -f(x,y,z)$$

which means

$$I = -I implies I = 0$$

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Math Genius: Finding $iiint_K (x^2 -z^2), dx, dy ,dz$

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Calculate the integral
$$
iiint_K (x^2 -z^2), dx, dy ,dz,,
$$

where $$K=left{(x,y,z):x,y,zge 0,x+y+zle 1right}$$

I have tried solving it but I really don’t understand how to handle the upper limits. I really need some guidance.

Update:
I got it to zero by using it’s symmetry as given in comments. Thanks for the help. I also confirmed it by calculating it myself with the values inputted.

The region has an $xz$ exchange symmetry, i.e.

$$(x,y,z) in K implies (z,y,x) in K$$

In other words

$$I = iiint_K f(x,y,z):dV = iiint_K f(z,y,x):dV$$

but in this problem for $f(x,y,z) = x^2-z^2$ we have that

$$f(z,y,x) = -f(x,y,z)$$

which means

$$I = -I implies I = 0$$

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Math Genius: Show that the tangent plane of the saddle surface $z=xy$ at any point intersects the surface in a pair of lines.

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My attempt: Let $f(x,y,z)=xy-z$ , (a,b,c)$in$the saddle surface, and calculate the total derivative $Df(a,b,c)=(b,a,-1)$ Then the tangent plane is $$g(x,y,z)=f(a,b,c)+Df(a,b,c)(x-a,y-b,c-z)=bx+ay-z+ab$$
Set $g(x,y,z)=f(x,y,z)$ to get the intersection got $bx+ay-xy+ab=0$.
I know that the equation can be written as $$bx+ay-xy+ab=bx-ay-z-c=0$$ But I have no idea to get a pair of lines which intersects the saddle surface.

First of all you made an error in the calculation:
$$begin{align}
g(x,y,z)=f(a,b,c)+Df(a,b,c)(x-a,y-b,color{red}{z-c})&=bx+ay-zcolor{red}{+c-2ab}\
&=bx+ay-zcolor{red}{-ab}
end{align}
$$

where we used $ab-c=0$.

Now the intersection of the surfaces can be found from the equation:
$$
xy-z=bx+ay-z-abimplies (x-a)(y-b)=0,
$$

which solutions are $x=a$ and $y=b$.

Substituting the values into equation of any of two surfaces one obtains that the intersection lines are:
$$
begin {cases}x-a=0\
ay-z=0
end {cases}quadtext {and}quad
begin {cases}y-b=0\
bx-z=0
end {cases}.
$$

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Math Genius: If $lvert z rvert < 1$ then $z^n to 0$ as $n to infty$

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$renewcommand{abs}[1]{lvert #1 rvert}$Let z be a complex number. I would like to show that if $abs{z} < 1$, then $z^n to 0$ as $ntoinfty$. Could anyone provide me with a proof of this? I have no trouble showing the corresponding statement if $z$ were a real number.

$|z^{n}|=|z|^{n} to 0$ by the real case since $|z|$ is a real number in $[0,1)$. If $epsilon >0$ then there exists $n_0$ such that $|z|^{n} <epsilon$ for all $n >n_0$. Hence $|z^{n}| <epsilon$ for all $n >n_0$. This is what it means to say that $z^{n} to 0$.

Let $z=re^{itheta}$. Then $z^n=r^ne^{itheta n}$. As $|e^{itheta n}|=1$, and $rlt1$, then $z^nto0$ as $ntoinfty$.

Let z=modz e^i0 z^n=modz^n e^i0*n as e^i0n is some complex no. And mod z^n is 0 so 0 into any no. Is anyways zero

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Math Genius: If $lvert z rvert < 1$ then $z^n to 0$ as $n to infty$

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$renewcommand{abs}[1]{lvert #1 rvert}$Let z be a complex number. I would like to show that if $abs{z} < 1$, then $z^n to 0$ as $ntoinfty$. Could anyone provide me with a proof of this? I have no trouble showing the corresponding statement if $z$ were a real number.

$|z^{n}|=|z|^{n} to 0$ by the real case since $|z|$ is a real number in $[0,1)$. If $epsilon >0$ then there exists $n_0$ such that $|z|^{n} <epsilon$ for all $n >n_0$. Hence $|z^{n}| <epsilon$ for all $n >n_0$. This is what it means to say that $z^{n} to 0$.

Let $z=re^{itheta}$. Then $z^n=r^ne^{itheta n}$. As $|e^{itheta n}|=1$, and $rlt1$, then $z^nto0$ as $ntoinfty$.

Let z=modz e^i0 z^n=modz^n e^i0*n as e^i0n is some complex no. And mod z^n is 0 so 0 into any no. Is anyways zero

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Math Genius: Show that the tangent plane of the cone $z^2=x^2+y^2$ at (a,b,c)$ne$0 intersects the cone in a line

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My attempt: Let $f(x,y,z)=x^2+y^2-z^2$ and calculate the total derivative $Df(a,b,c)=(2a,2b,-2c)$. The tangent plane is $$g(x,y,z)=f(a,b,c)+Df(a,b,c)(x-a,y-b,z-c)=2ax+2by-2cz-a^2-b^2+c^2$$ Then let $g(x,y,z)=f(x,y,z)$ to find the intersction got $$(x-a)^2+(y-b)^2-(z-c)^2=0$$ Does this equation means the tangent plane intersects the cone in a line?

You took a point $(a,b,c)$ belonging to the cone $C equiv z^2=x^2+y^2$. So you must have $a^2+b^2=c^2$ and the equation of the tangent plane simplifies to $2ax+2by-2cz=0$.

From there, it is easy to see that for all points $P_lambda = (lambda a, lambda b, lambda c)$ that belongs to a line included in $C$, the equation of the tangent plane at $P_lambda$ is also $2ax+2by-2cz=0$. That proves the expected resut.

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Math Genius: Show that the tangent plane of the cone $z^2=x^2+y^2$ at (a,b,c)$ne$0 intersects the cone in a line

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My attempt: Let $f(x,y,z)=x^2+y^2-z^2$ and calculate the total derivative $Df(a,b,c)=(2a,2b,-2c)$. The tangent plane is $$g(x,y,z)=f(a,b,c)+Df(a,b,c)(x-a,y-b,z-c)=2ax+2by-2cz-a^2-b^2+c^2$$ Then let $g(x,y,z)=f(x,y,z)$ to find the intersction got $$(x-a)^2+(y-b)^2-(z-c)^2=0$$ Does this equation means the tangent plane intersects the cone in a line?

You took a point $(a,b,c)$ belonging to the cone $C equiv z^2=x^2+y^2$. So you must have $a^2+b^2=c^2$ and the equation of the tangent plane simplifies to $2ax+2by-2cz=0$.

From there, it is easy to see that for all points $P_lambda = (lambda a, lambda b, lambda c)$ that belongs to a line included in $C$, the equation of the tangent plane at $P_lambda$ is also $2ax+2by-2cz=0$. That proves the expected resut.

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Math Genius: Converting an Integral to Spherical Coordinates

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I need to convert the following integral to spherical coordinates

$displaystyle int_{-1}^{1} int_{0}^{sqrt{1-x^2}}int_{0}^{x^2 + y^2} y^2 dz dy dx$

My main issue is with limits of $z$.

Using limits of $x$ and $y$, I know we need to consider the upper half of the circle $x^2 + y^2 =1$

Now, $z = x^2 + y^2$ is a paraboloid opening upwards,cut off by plane $z =1$

So, by this logic the limits for $z$ should be:

$x^2 + y^2 le z le 1$,

I don’t get how the limits of $z$ are between $ 0$ and $x^2 + y^2$,can someone please clear this confusion for me ?

Thank you.

Sketch for solution: as the integral is defined you have that
$$
0leqslant zleqslant x^2+y^2,quad 0leqslant y^2leqslant 1-x^2,quad 0leqslant x^2leqslant 1tag1
$$

The spherical coordinates are given by
$$
x:=rcos alpha sin beta ,quad y:=r sin alpha sin beta ,quad z:=rcos beta \
text{ for }alpha in [0,2pi ),quad beta in [0,pi ),quad rin [0,infty )tag2
$$

Therefore $(1)$ becomes
$$
0leqslant rcos beta leqslant r^2sin^2 beta ,quad 0leqslant r^2sin^2 beta leqslant 1,quad 0leqslant r^2cos^2 alpha sin^2 beta leqslant 1tag3
$$

what simplifies to
$$
0leqslant rcos beta leqslant r^2sin ^2beta leqslant 1tag4
$$

What remains is to find the range of valid values for $r, alpha $ and $beta $ from $(4)$ and it defining bounds (stated on $(2)$), and rewrite the integral accordingly.

The limits of the rightmost integral are $0$ and $x^2+y^2$. Clearly $0le x^2+y^2$ for any real $x$ and $y$. Therefore the limits are well-defined and $0le zle x^2+y^2$. The lateral surface of the body are the plane $y=0$ and the cylinder $x^2+y^2=1$ (for $yge0$).

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