surface area formulas of all 3D shapes?

Answer 1

http://argyll.epsb.ca/jreed/math9/strand3/3107.htm
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Answer 2

Surface area is the area of a given surface. Roughly speaking, it is the "amount" of a surface (i.e., it is proportional to the amount of paint needed to cover it), and has units of distance squared. Surface area is commonly denoted S for a surface in three dimensions, or A for a region of the plane (in which case it is simply called "the" area).

The following tables gives lateral surface areas S for some common surfaces. Here, r denotes the radius, h the height, e the ellipticity of a spheroid, p the base perimeter, s the slant height, a the tube radius of a torus, and c the radius from the rotation axis of the torus to the center of the tube. Note that many of these surfaces are surfaces of revolution, for which Pappus's centroid theorem can often be used to easily compute the surface area.

surfaceS
conepirsqrt(r^2+h^2)
conical frustumpi(R_1+R_2)sqrt((R_1-R_2)^2+h^2)
cube6a^2
cylinder2pirh
oblate spheroid2pia^2+(pic^2)/eln((1+e)/(1-e))
prolate spheroid2pia^2+(2piac)/esin^(-1)e
pyramid1/2ps
pyramidal frustum1/2ps
sphere4pir^2
spherical lune2r^2theta
torus4pi^2ac
zone2pirh

Even simple surfaces can display surprisingly counterintuitive properties. For instance, the surface of revolution of y==1/x around the x-axis for x>=1 is called Gabriel's horn, and has finite volume but infinite surface area.

For many symmetrical solids, the interesting relationship
S==(dV)/(dr)
(1)

holds between the surface area S, volume V, and inradius r. This relationship can be generalized for an arbitrary convex polytope by defining the harmonic parameter h in place of the inradius r (Fjelstad and Ginchev 2003).

If the surface is parameterized using u and v, then
S==int_S|T_uxT_v|dudv,
(2)

where T_u and T_v are tangent vectors and axb is the cross product. If z==f(x,y) is defined over a region R, then
S==intint_(R)sqrt(((partialz)/(partialx))^2+((partialz)/(partialy))^2+1)dA,
(3)

where the integral is taken over the entire surface (Kaplan 1992, 3rd ed. pp. 245-248).

Writing x==x(u,v), y==y(u,v), and z==z(u,v) then gives the symmetrical form
S==intint_(R^')sqrt(EG-F^2)dudv,
(4)

where R^' is the transformation of R, and
E=((partialx)/(partialu))^2+((partialy)/(partialu))^2+((partialz)/(partialu))^2
(5)
F=(partialx)/(partialu)(partialx)/(partialv)+(partialy)/(partialu)(partialy)/(partialv)+(partialz)/(partialu)(partialz)/(partialv)
(6)
G=((partialx)/(partialv))^2+((partialy)/(partialv))^2+((partialz)/(partialv))^2
(7)

are coefficients of the first fundamental form

s

Answer 3

the exterior portion of an oblong stable is expressed by utilising the formulation: A = 2(wh+wl+lh) so which you have 3 values to calculate: the fabricated from the width and the top (wh), the fabricated from the width and length (wl), and the fabricated from the dimensions and top (lh) wh = 6 one million/2 * 10 = sixty 5 sq. yards start up by utilising changing the width to its fractional sort. to transform 6 one million/2 to fractional sort, multiply the entire sort (6) by utilising the denominator of the fraction (2) and upload the numerator (one million). 6 situations 2 is 12, upload one million is thirteen. So the fractional form of 6 one million/2 = thirteen/2. Now take that results of thirteen/2 and multiply by utilising 10 (the top). you get one hundred thirty/2. Divide that out and you get sixty 5. that's the portion of the rectangle formed by utilising the width and the top. Following a similar technique, you will get: wl = 6 one million/2 * 5 one million/2 = thirteen/2 * 11/2 = (thirteen*11)/(2*2) = 143/4 = 35 3/4 sq. yards and lh = 5 one million/2 * 10 = 11/2 * 10 = fifty 5 sq. yards now you have each and all the phrases referred to as for interior the equation above: A = 2(wh+wl+lh) = 2(sixty 5+35 3/4+fifty 5) = 2(a hundred and fifty five 3/4) = 310 + 6/4 = 311 one million/2 sq. yards

Answer 4

this website should give you all

http://mathforum.org/dr.math/faq/formulas/index.html