## 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

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

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

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

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

## 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

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

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

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

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

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

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

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

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

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

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

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

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\$

$$renewcommand{abs}[1]{lvert #1 rvert}$$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 $$|z|^{n} for all $$n >n_0$$. Hence $$|z^{n}| 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

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

$$renewcommand{abs}[1]{lvert #1 rvert}$$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 $$|z|^{n} for all $$n >n_0$$. Hence $$|z^{n}| 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

## 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

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

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

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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