[PDF] Math 2400: Calculus III Introduction to Surface Integrals PDF Project10

roughly given by ∆S = rr × rθ∆r ∆θ = (2r cosθ,2r sinθ, r)∆r ∆θ = √ 5r∆r ∆θ √ 5r dr = √ 5π Since the surface is a cone, we can confirm our result using the formula for the lateralsurface area of a cone, S = πrs, where s is the slant height Here the radius is 1 and the slant height is √ 5, confirming our result

Use calculus to derive the formula for the volume of a cone of radius R and height h Another way to get the lateral surface area is to use spherical coordinates

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You can think of dS as the area of an infinitesimal piece of the surface S To define the integral (1) To do the integration, we use spherical coordinates ρ, φ, θ

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Triple integral in spherical coordinates Use spherical coordinates to express region between the sphere x2 + y2 + z2 = 1 The bottom surface is the cone:

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As the circle is rotated around the z-axis, the relationship stays the same, so ρ = 2 sinφ is the equation of the whole surface To determine the limits of integration,

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(a) Find the volume of an ice cream cone bounded by the cone z = √x2 + y2 and the (b) In spherical coordinates, the hemisphere is given by ρcos(φ) = √

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by hand, and then convert them to spherical coordinates A 2 2 2 16 4 ρ + In HW set #7 number 7 we found the volume of an ice-cream cone which was bounded Determine the surface area of the entire solid described in problem 4

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and above the cone given by φ = π/3 in spherical coordinates (5) E is where recall that the surface area element on a sphere of radius a is rφ × rθ = a2 sin φ

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To compute a surface integral over the cone, one needs to compute rθ × rz = ⟨− z sinθ, and it is obtained using spherical coordinates In this case, we have

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Math 2400: Calculus IIIIntroduction to Surface Integrals - Generalizing the formula for surface areaYou recently learned how to nd the area of a surface by parameterizing, then evaluating the appropriate integral.

The rst exercise is a review of that concept. In the second problem we will generalize the idea of surface area,

introducing a new type of integral: surface integrals of scalar elds. 1. Find the surf acearea of th epart of the surface z2= 4x2+ 4y2lying betweenz= 0 andz= 2. (a) Find the in tersectionof the surface z2= 4x2+ 4y2andz= 2. (b) Graph the s urfacew eare trying to nd th earea of. (c)

P arameterizeth esurface using cylindrical co ordinatesto get r(r;). Don't forget to include the intervals

for the parameters. (d) A rectangle with d imensions rand gets mapped by the parameterization onto the surface, to a

patch that is roughly a parallelogram. The area of this patch is approximately calculated by the formula

S=jrrrjr. Calculate the area Sfor this surface.

(e)

Find the total surface area b yin tegrating.

Solution:

The conez2= 4x2+ 4y2intersects the planez= 2 when 22= 4x2+ 4y2, orx2+y2= 1. This is a circle of radius 1, at an altitude ofz= 2. Since z0, we are dealing with only the upper nappe of the conez2= 4x2+4y2. Thex-ztrace of the upper nappe of the conez2= 4x2+ 4y2has formula z

2= 4x2, orz= 2jxj. So it is narrower than a right-circular cone.

To parameterize the surface using cylindrical coordinates, notice that the top view of the surface is a disc of

radius 1. To ll this disc, our parameters range from 0r1 and 02. Noting that for the upper nappe of the cone,z= 2px

2+y2, our parameterization is given byr(r;) = (rcos;rsin;2r).

Dierentiating givesrr= (cos;sin;2) andr= (rsin;rcos;0). The area of a patch on the surface is roughly given by S=jrrrjr=j(2rcos;2rsin;r)jr=p5rr.

Integrating gives that the area of the cone is

Z 2 =0Z 1 r=0p5rdrd= 2Z 1

0p5rdr=p5:

Since the surface is a cone, we can conrm our result using the formula for the lateralsurface area of a cone,

S=rs, wheresis the slant height. Here the radius is 1 and the slant height isp5, conrming our result.

2. In an in tegralfor surface area, w euse the dieren tialjrurvjdudv, which geometrically represents: Solution:the area of the tiny paralellogram which the parameterization maps adu-by-dvrectangle onto. 3.

A studen tasks \W eha vecalculated

RR Rjrurvjdudv, which is a double integral. I thought a double- integral gives a volume, so why does this integral only give an area?" Respond to this student.

Solution:In the double integralRR

Rf(x;y)dA, the dierentialdArepresents the area of a tiny rectangle.

Iff(x;y) is a height, thenf(x;y)dAgives a volume of a rectangular box with a very tiny base. The integral

then gives a volume. In our integral, all of the expressionjrurvjdudvrepresents an areadA. Thus there

is no height in the integrand. Our integral sums areas, not volumes, so our nal result represents an area,

not a volume.

Math 2400: Calculus IIIIntroduction to Surface Integrals - Generalizing the formula for surface areaWe have seen that the area of a parameterized surfacer(u;v) over the regionRcan be found by calculating

ZZ R

1jrurvjdudv:

The idea of a surface integral is to generalize by replacing the \1" with an arbitrary function. 4.

Supp osethe surface of p roblem1 has a v ariabledensit yof (x;y;z) =p4z2. Find its total mass. Assume

the units of mass are grams, and the units of distance are meters. (a)

What are the units of ?

(b) In the pr eviousproblem, w ecalculated the area of a patc has jrrrjr=p5rr. (c)

What is the mass of a patc h?

(d)

In tegrateto nd the total mass.

Solution:The densityis a mass per unit area, so in this case it is given in grams per square meter.

The mass of small patch on the surface is given by the density at that location times the area of the

patch. This gives m=p4z2p5rr: Along the surface of the cone,z2= 4x2+ 4y2= 4r2. Substituting gives m=p5 p44r2rr: Using the same parameterization as in the previous problem, we calculate the total mass: m=Z 2 =0Z 1 r=0p5 p44r2rdrd = 2 p5 Z 2 =0Z 1 r=0p1r2rdrd = 4 p5Z 1 r=0p1r2rdr=4p53 To do a plausibility-check on this result, recall that the area of the surface is p5. This means that the average density of the surface must be 43
. The density of the cone decreases from= 2 at the bottom point of the cone to= 0 at the very top. The density atz= 1, halfway up the cone, is=sqrt3,

which is larger than what we calculated as the average density of the cone. This makes sense since more

of the cone lies above that halfway point, and the upper portion is less dense than the lower portion.

The surface integral of a functionf(x;y;z) (i.e., a scalar eld) over a surfaceSis writtenZZ S f(x;y;z)dS. It is computed byparameterizing the surfaceSasr(u;v) and computingZZ R

f(r(u;v))jrurvjdA, (whereRdenes the domain ofuandv).5.Practice at home: Re-do the in tegralsin problems 1 and 4 u singspherical co ordinates.Mak esure y ouget

the same numerical answers as when you did the calculations in cylindrical coordinates. 6.

Practice at home: Use spherical co ordinatesto n dthe area of the part of the sphere x2+y2+z2= 16 that

lies above the planez= 2. Then nd the mass of that surface if the density is given by=z.quotesdbs_dbs5.pdfusesText_10

[PDF] [PDF] Math 2400: Calculus III Introduction to Surface Integrals