[PDF] Lecture 22, November 23 • Surface integrals

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

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

[PDF] Area and Volume Problems

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

[PDF] Section 74 - Area of a Surface Problem 1 (Exercise 7410) Find the

picture of an ice cream cone (from the cone) with a scoop of ice cream on top all our points to lie on the unit sphere, so spherical coordinates are probably our

[PDF] Surface Integrals - 1802 Supplementary Notes Arthur Mattuck

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

[PDF] Integrals in cylindrical, spherical coordinates - MSU Math

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

[PDF] Limits in Spherical Coordinates - MIT OpenCourseWare

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,

[PDF] Solutions

(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) = √

[PDF] Mat 241 Homework Set 10

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

[PDF] Math 120: Practice for the final

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

[PDF] Lecture 22, November 23 • Surface integrals - TCD Maths home

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

[PDF] confidence interval calculator

[PDF] configuration électronique du carbone

[PDF] configuration profile reference

[PDF] configuration web_url

[PDF] confinement france date du début

[PDF] confinement france voyage à l'étranger

[PDF] confinement france zone rouge

[PDF] congo two letter country code

[PDF] congressional research service f 35

[PDF] conjonctivite remède de grand mère

[PDF] conjugaison exercices pdf

[PDF] conjunction words academic writing

[PDF] connecteur de cause et conséquence

[PDF] connecteur logique juridique pdf

[PDF] connecteurs logiques d'introduction

Lecture 22, November 23

Surface integrals.The integral off(x,y,z) over a surfaceσinR3is f(x,y,z)dS=∫∫ f(x(u,v),y(u,v),z(u,v))· ||ru×rv||dudv, wherer(u,v) =⟨x(u,v),y(u,v),z(u,v)⟩is the parametric equation of the surface.

1 +f2x+f2ydxdy.

Example 1 (Cylinder).

The parametric equation of the cylinderx2+y2= 1 is and it is obtained using cylindrical coordinates. In this case, we have r θ×rz=⟨-sinθ,cosθ,0⟩ × ⟨0,0,1⟩=⟨cosθ,sinθ,0⟩, cos

2θ+ sin2θ= 1

and one can use these facts to compute any surface integral over the cylinder.

Example 2 (Cone).

2+y2is

x To compute a surface integral over the cone, one needs to compute r

θ×rz=⟨-zsinθ,zcosθ,0⟩ × ⟨cosθ,sinθ,1⟩=⟨zcosθ,zsinθ,-z⟩,

z 2.

Example 3 (Sphere).

The parametric equation of the spherex2+y2+z2= 1 is and it is obtained using spherical coordinates. In this case, we have r

θ×rϕ=⟨-sinθsinϕ,cosθsinϕ,0⟩ × ⟨cosθcosϕ,sinθcosϕ,-sinϕ⟩

and the fact that||r||= 1 implies that||rθ×rϕ||= sinϕ.

Example 4 (A general example).

The graph ofz=f(x,y) can be described by

To compute a surface integral over this graph, one needs to compute r x×ry=⟨1,0,fx⟩ × ⟨0,1,fy⟩=⟨-fx,-fy,1⟩,

1 +f2x+f2y.

Example 5.

We compute the integral

σz2dSin the case thatσis the part of the

2+y2that lies betweenz= 0 andz= 1. As in Example 2, we have

2dz dθ, so the given integral becomes

z2dS=∫ 2π

0∫

1 0

2dz dθ=∫

2π 2 4 2 2

Example 6.

Consider the lamina that occupies the part of the paraboloidz=x2+y2 that lies below the planez= 1. If its density is given byδ(x,y,z), then its mass is

[PDF] [PDF] Lecture 22, November 23 • Surface integrals - TCD Maths home

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