Math Genius: Finding the Volume of a Torus by rotating a circle with equation \$x^2 + (y-R)^2 = r^2\$ a different way

I was trying to find the Volume of a Torus and after sketching it out in Desmos, I thought I found a solution but it doesn’t work (or at least when I evaluated it it didn’t work).

$$V = 2πint_{-r}^r(sqrt{r^2-x^2}+R)^2-R^2 dx$$

My reasoning was that I would take the integral of the top half of the circle subtracted by the integral of the line cutting the circle in half and then multiply the whole thing by two to get the total volume. Any help would be appreciated, thanks!

Tagged : / /

Math Genius: Volume of Solids of Revolution with Hyperbola

The area bounded above by the line $$y = 3$$, below by the line $$y = 0$$, on the left by the y-axis and on the right by an arc of the hyperbola $$9x^2 – 16y^2 = 144$$ is rotated around the x-axis. Find the volume of the solid generated.

I first found the intersection point between the hyperbola and the line $$y = 3$$ to be at $$x = 4sqrt{2}$$. From there, I first found the area from $$x = 0$$ to $$x = 4$$ and then from $$x = 4$$ to $$x = 4sqrt{2}$$. The following is my work:

From $$x = 0$$ to $$x = 4$$:

$$int_0^4 pi(3)^2 dx = 36pi$$

From $$x = 4$$ to $$x = 4sqrt{2}$$:

$$int_4^{4sqrt{2}} biggl(frac{9x^2}{16} – 9biggl)pi dx = 24pibiggl(1 – frac{sqrt{2}}{2}biggl)$$

I then added both of these areas. However, in the book where I was reading this problem, the answer is supposed to be $$24pibiggl(2sqrt{2} – 1biggl)$$. I’m new to solids of revolution and so far I’ve only learned how to find the volume using the cylindrical method.

If anyone can help me understand what I did wrong or give a hint on how to solve the problem I would appreciate it.

Since $$y=3$$ lies above the branch of the hyperbola, your second integral should be
$$int_4^{4sqrt{2}} left(9-frac{9x^2-144}{16}right) dx = 48 sqrt{2} pi -60 pi.$$

It is more convenient to integrate with the cylinder method

$$int_0^3 2pi yx(y)dy =frac23 pi int_0^3 ysqrt{144+16y^2}dy=24pi(2sqrt2-1)$$

Math Genius: Double/Volume integral in parametric form

Find the volume of the tetrahedron with vertices at $$(0,0,0),(a,0,0),(0,b,0),(0,0,c)$$

The most straight forward approach would be evaluating the following integral in Cartesian coordinates system.
$$V=int_0^aint^{b(1-x/a)}_{0}int^{c(1-x/a-y/b)}_0dzdydx=frac{abc}{6}$$
(which I can make sense of.)

But I tried two alternative methods each of which involves parametric equations:

Describing the surface that joins the three vertices, with a vector equation, I got
$$vec{r}=(a,0,0)+lambda(-a,b,0)+mu(-a,0,c)=(a(1-lambda-mu),lambda b,mu c)=(x,y,z) tag{1}$$
(here I used $$vec{AB}=(-a,b,0)$$ and $$vec{AC}=(-a,0,c)$$ as the two directional vectors.)

and
$$x=a(1-lambda-mu),y=lambda b, z=mu c tag{2}$$
as the three parametric equations that describe the surface, where $$0leq yleq b$$ and $$0leq zleq c$$ give the limits $$0leq lambda leq 1$$ and $$0leq mu leq 1$$.

Taking $$dV=xdydz$$:
$$dV=x(lambda,mu)frac{dy}{dlambda}frac{dz}{dmu}dlambda dmu=abc(1-lambda-mu)dlambda dmu tag{3}$$
Integration yields $$V=abcint^{mu=1}_{mu=0}int^{lambda=1}_{lambda=0}(1-lambda-mu)dlambda dmu=0 tag{4}$$
which is clearly wrong.

Alternatively, by taking $$dV=dxdydz$$, I got some rather dodgy differential terms:
$$dV=left(frac{partial x}{partial lambda}dlambda+frac{partial x}{partial mu}dmuright)frac{dy}{dlambda}frac{dz}{dmu}dlambda dmu=-abc(dlambda+dmu)dlambda dmu tag{5}$$
which has terms like $$dlambda^2$$ (the square of a differential is not something I’ve seen before, except for $$ds^2=g_{munu}dx^{mu}dx^{nu}$$ but that’s a different story…).

Can someone please explain where my conceptual errors lie?

Or is there a correct approach involving parametric equations?

Math Genius: Integral of product of 3D Gaussians

I am reading a paper, Keep it SMPL: Automatic Estimation of 3D Human Pose and Shape from a Single Image, CVPR 2016, that models the human parts with capsules for estimating the interpenetration penalty and then abstracts that with 3D Gaussians which corresponds to the bones (for calculating the intersection of parts estimated with 3D spheres because calculating the volume of intersecting capsules is very much challenging).

There’s this formula in the paper that I can’t wrap my head around it.

In this formula, that is for integral of the product of Gaussians corresponding to incompatible part, specifically, I don’t understand the following part (I understand what C_i(theta, beta), C_j(theta, beta), sigma_i^2, and sigma_j^2 are):

I understand that product of two Gaussians does have exp as well as sigma_i^2 + sigma_j^2 in the denominator of exp. However, I don’t understand why we are using the norm 2 squared of the difference of centers in the nominator of exponential function??

Below, I have added screenshots from the paper that goes over the formula (however, mathematically, it is still not clear to me how we arrived at this formula)

I reformulated your kernel as follows:
$$E(theta;beta)=sum_isum_j exp(frac{(C_i – C_j)^p}{sigma^2_i+sigma^2_j})$$
The smoothness parameter $$p$$ affects the correlation trend between $$C_i$$ and $$C_j$$. Suppose we test $$p=0.1, 1, 2$$. So $$p=0.1$$ makes the correlation to be near discontinuity between $$C_i$$ and $$C_j$$. For $$p=1$$, we have a non-differentiable kernel which might be useful for non-differentiable functions, and for $$p=2$$ we have a smooth basis function which is differentiable for all orders. Although you fix the parameter $$p$$ for all of your dimensions, it can be estimated for every dimension. For $$p=2$$, we have a squared exponent function.

Tagged : / / / /

Server Bug Fix: Find the volume between the surface \$x^2+y^2+z=1\$ and \$ z=x^2+(y-1)^2\$

I’m trying to find the volume between the surface $$x^2+y^2+z=1$$ and $$z=x^2+(y-1)^2$$ but nothing works for me.

I made the plot and it looks like this:

How could you start? Any recommendation?

Try checking where the two surfaces intersect. Solve the given equations for $$z$$ and set them equal:

begin{align*} 1-x^2-y^2&=x^2+(y-1)^2\ 0&=2x^2+2y^2-2y\ 0&=x^2+y^2-y\ 0&=x^2+left(y-frac12right)^2-frac14\ x^2+left(y-frac12right)^2&=frac1{2^2} end{align*}

which corresponds to the cylinder of radius $$frac12$$ with cross sections parallel to the $$(x,y)$$ plane and centered over the point $$left(0,frac12,0right)$$.

With this in mind, a change to cylindrical coordinates is the “obvious” move. Take

$$begin{cases}x=rcostheta\y=frac12+rsintheta\z=zend{cases}impliesmathrm dx,mathrm dy,mathrm dz=r,mathrm dr,mathrm dtheta,mathrm dz$$

The solid (call it $$S$$) is then given by the set

$$S=left{(r,theta,z)mid0le rlefrac12,0lethetale2pi,r^2-rsintheta+frac14le zlefrac34-r^2-rsinthetaright}$$

where the last inequality follows from

begin{align*} &x^2+(y-1)^2le zle1-x^2-y^2\ &implies r^2cos^2theta+left(rsintheta-frac12right)^2le zle1-r^2cos^2theta-left(frac12+rsinthetaright)^2\ &implies r^2-rsintheta+frac14le zlefrac34-r^2-rsintheta end{align*}

The volume is then

$$iiint_Smathrm dx,mathrm dy,mathrm dz=int_0^{2pi}int_0^{frac12}int_{r^2-rsintheta+frac14}^{frac34-r^2-rsintheta}r,mathrm dz,mathrm dr,mathrm dtheta$$

$$=fracpi{16}$$

The height $$h$$ of the region at a given $$x,,y$$ is$$h=1-2x^2-y^2-(y-1)^2=2(1/4-x^2-(y-1/2)^2),$$and the relevant $$x,,y$$ are those for which this is $$ge0$$. Parameterizing such $$x,,y$$ by $$x=rhocostheta,,y=tfrac12+rhosintheta$$ with $$0lerholetfrac12,,0lethetale2pi$$ gives the volume element as$$hdxdy=hrho drho dtheta=2(1/4-rho^2)rho drho dtheta,$$so the volume is$$4piint_0^{1/2}rho(1/4-rho^2)drho=-pi[(1/4-rho^2)^2]_0^{1/2}=frac{pi}{16},$$in agreement with @user170231’s answer in terms of a triple integral.

Here is how I would do it.

1. Solve for $$z$$ in both and set equal to each other, to get the intersection curve in the $$xy$$-plane.
2. Integrate the difference between $$z$$‘s over the inside of that curve to get $$Delta V$$.

Math Genius: Finding Volume of a solid via integration

I’m new to integration and I have this question I have to work out. Please let me know if I went wrong. Thank you.

I have to find the volume of a shape by the bounds below, about the y-axis:
$$y = ln(x), y=1, y=2, x=0$$

Which on a graph looks as so:

With an area of a circle surface :
$$A= pi r^2$$
and since it is about the y-axis, $$r = x space dy$$

With $$y= ln(x)$$, $$space x = e^y$$

So I established this integral and worked out a volume:
$$int_1^2(pi x^2 )dy$$
$$piint_1^2(e^{2x})dy$$
$$frac12 piint_1^2(e^{u})du$$
$$frac12 piint_1^2(e^{2x})$$
$$frac12pi[e^{4}-e^{2}]$$
$$V approx 74.155…$$

Tagged : /

Math Genius: Triple integral and mass

An object occupies the solid region bounded by the upper nappe of the cone $$z^2=9x^2+y^2$$ and the plane $$z=9$$ Find the total mass of the object if the mass density at (x,y,z) is equal to the distance from (x,y,z) to the top.

I switched to spherical coordinates, I can’t find the right bounds… The answer is $$243/4 pi$$, I tried many ways, nothing is working so I would appreciate if someone showed me the correct way to do this.

The density is $$rho(x,y,z) = 9-z$$, the mass is $$M = iiint_V rho(x,y,z) dxdydz$$, where $$V$$ is the region occupied by the solid in your question.

$$M = int_0^9 dz(9-z)iint_{Omega_z} dx dy$$
where $$Omega_z = { (x,y): 0 leq 9x^2+y^2leq z^2}$$ and $$|Omega_z| = iint_{Omega_z} dx dy= frac{pi}{3}z^2$$

Hence $$M = int_0^9 (9-z) frac{pi}{3}z^2 dz = frac{729 pi}{4}$$

Tagged : / /

Math Genius: find the volume using the method of disks or washers via an integral

The volume of the solid obtained by rotating the region enclosed by $$y=frac{1}{x^2} , y=0, x=3, x=8$$ about the line $$y=-1$$.

How do I find the volume? I need help.

I tried
$$piint_8^3(frac{1}{x^2}-(-1))^2dx$$ but i did not find the correct answer

The disk integral is

$$piint^8_3[ (frac{1}{x^2}-(-1))^2-(0-(-1))^2]dx = pileft( -frac1{3x^3}-frac2xright)bigg|_3^8$$

Tagged : / /

Math Genius: 3-dimensional surface volumn of a 4-dimensional hyperellipsoid

The circumference of a 2-dimensional ellipse and the surface area of a 3-dimensional ellipsoid can be given in close-forms using the Legendre’s elliptic integrals as
$$C_2=4aE(sqrt{1-b^2/a^2})$$
$$A_3=2pileft[c^2+frac{bc^2}{sqrt{a^2-c^2}}Fleft(arcsinsqrt{1-c^2/a^2},frac{asqrt{b^2-c^2}}{bsqrt{a^2-c^2}}right)+bsqrt{a^2-c^2}Eleft(arcsinsqrt{1-c^2/a^2},frac{asqrt{b^2-c^2}}{bsqrt{a^2-c^2}}right)right]$$
where $$a$$, $$b$$ and $$c$$ are positive real numbers representing the semi-axes (radii) and $$E(k)$$, $$F(phi,k)$$ and $$E(phi,k)$$ are the complete elliptic integral of the second kind and the incomplete elliptic integral of the first kind and the second kind respectively.

However, I have troubles when I attempt to calculate a formula for the surface volume of a 4-dimensional hyperellipsoid. Let $$a$$, $$b$$, $$c$$ and $$d$$ to be the semi-axes of the hyperellipsoid such that
$$frac{x_1^2}{a^2}+frac{x_2^2}{b^2}+frac{x_3^2}{c^2}+frac{x_4^2}{d^2}=1$$
then the surface volume can be given by the integral
$$V_4=16intlimits_{t_1=0}^aintlimits_{t_2=0}^{bsqrt{1-t_1^2/a^2}}intlimits_{t_3=0}^{csqrt{1-t_1^2/a^2-t_2^2/b^2}}sqrt{frac{1-(1-d^2/a^2)t_1^2/a^2-(1-d^2/b^2)t_2^2/b^2-(1-d^2/c^2)t_3^2/c^2}{1-t_1^2/a^2-t_2^2/b^2-t_3^2/c^2}}dt_3dt_2dt_1$$
that I cannot integrate. How can one integrate this integral to a close-form? Or, alternately, is there another way to calculate a formula for the surface volum of a 4-dimensional hyperellipsoid?

Edit: The integral can be reduced to
$$V_4=frac{16abcpi}{3}intlimits_0^1t^2left(frac{d^2/a^2}{1-(1-d^2/a^2)(1-t^2)^2}+frac{d^2/b^2}{1-(1-d^2/b^2)(1-t^2)^2}+frac{d^2/c^2}{1-(1-d^2/c^2)(1-t^2)^2}right)sqrt{frac{2-t^2}{(1-(1-d^2/a^2)(1-t^2)^2)(1-(1-d^2/b^2)(1-t^2)^2)(1-(1-d^2/c^2)(1-t^2)^2)}}dt$$
but I still cannot solve it.

Math Genius: Minimum surface of a cone

Imagine, I choose \$a\$ as the volume of the cone. Find an expression with \$a\$ as variabele for the minimum surface.

My Work:

• \$\$text{Volume}=frac{pi r^2h}{3}=ato r=sqrt{frac{3a}{pi h}}\$\$
• \$\$text{Surface Cone}=pi r^2+pi rsqrt{r^2+h^2}\$\$
• Filling in my \$r\$ in my surface expression:
\$\$text{Surface Cone}=pileft[sqrt{frac{3a}{pi h}}right]^2+pi left[sqrt{frac{3a}{pi h}}right]sqrt{left[sqrt{frac{3a}{pi h}}right]^2+h^2}=frac{3a}{h}+sqrt{3}sqrt{frac{a}{h}}sqrt{frac{3a}{h}+h^2pi}\$\$
• Take the derivative, to find the minumum:
\$\$frac{text{d}}{text{d}h}text{Surface Cone}=0Longleftrightarrowspacespacespace h=2sqrt[3]{a}sqrt[3]{frac{3}{pi}}\$\$
• Substitute the found \$h\$ in the \$text{Surface Cone}\$ expression:

\$\$text{Minimum Surface Cone}=2sqrt[3]{3^2}sqrt[3]{pi}sqrt[3]{a^2}\$\$

Am I right?

Let \$V=frac{1}{3}pi r^2h\$ and we have to minimize \$\$A(r,h)=pi r^2+pi rsqrt{r^2+h^2}\$\$ we have \$\$h=frac{3V}{pi r^2}\$\$ and our objective function will be
\$\$A(r,frac{3V}{pi r^2})=pi r^2+sqrt{r^2+frac{9V^2}{pi^2 r^4}}\$\$ this must be differentiated with respect to \$r\$.

Tagged :