Math 225: Calculus III   Solution 8

Initialization

<<Calculus`Math225`

i = {1, 0, 0}; j = {0, 1, 0}; k = {0, 0, 1};

Problem 1

Evaluate the integral
Integrate[Cos[x^2 + y^2], {x,-3,3}, {y,0,Sqrt[9-x^2]}]
by changing to polar coordinates. (Recall that in Mathematica the order of integration is right to left so that in this example the integral with respect to y is done first and then the integral with respect to x.)

Solution

The limits of integration define the region by the inequalities
    0 <= y <= Sqrt[9-x^2]
   -3 <= x <= 3

This is a semicircle of radius 3 which is described in polar coordinates by
    0 <= r <= 3
    0 <= t <= Pi
The function Cos[x^2 + y^2] becomes Cos[r^2], and we must add a factor of r to the integral for the area element r dr dt.

Integrate[ Cos[r^2] r, {r,0,3}, {theta,0,Pi}]

Problem 2

Find the centroid of the plane region that lies below the curve r == Sqrt[1+theta] and above the x-axis. (For this problem it is best to use NIntegrate[]).

Solution

First, let us draw a plot of the spiral r == Sqrt[1+theta] (this is not required). We'll switch to t instead of theta to save space.

region=ParametricPlot[
{Sqrt[1+t] Cos[t], Sqrt[1+t] Sin[t]},{t,0,Pi}]

The coordinates of the centroid, {xbar, ybar}, are the average values of
x = r Cos[t] and y = r Sin[t].

area = Integrate[1*r, {t,0,Pi}, {r,0,Sqrt[1+t]}]

xbar = NIntegrate[r Cos[t]*r,{t,0,Pi},
{r,0,Sqrt[1+t]}]/N[area]

xbar = NIntegrate[r Cos[t]*r,{t,0,Pi},
{r,0,Sqrt[1+t]}]/N[area]

ybar = NIntegrate[r Sin[t]*r,{t,0,Pi},
{r,0,Sqrt[1+t]}]/N[area]

Let's see where this point lies in the region (again, this is not required.)

Show[region,Graphics[
{PointSize[0.02],Point[{xbar,ybar}]}]]

Problem 3

Find the volume of the solid enclosed in the first octant by the parabolic sheets
y == 1 - x^2 and z == 1 - y^2.

Solution

Here's what the solid looks like (this is not required):

side=ParametricPlot3D[{x,1-x^2,t*(1-(1-x^2)^2)},
     {x,0,1},{t,0,1}];

top=ParametricPlot3D[{t Sqrt[1-y],y,1-y^2},{t,0,1},{y,0,1}];

Show[top,side,ViewPoint->{2.816, 1.415, 1.231}]

The top surface is z == 1-y^2, and the bottom surface is z == 0. These intersect in the first octant along the line, y == 1. The other equations define the region in the plane:
    0 <= y <= 1 - x^2
    0 <= x <= 1

volume = Integrate[1, {x,0,1},{y,0,1-x^2},{z,0,1-y^2}]

Problem 4

Find the volume of the solid bounded below by the plane 2y + z == -2 and above by the paraboloid x^2 + (y+1)^2 + z == 8.

Solution

The top surface is z == 8 - x^2 - (y+1)^2, and the bottom surface is
z == -2 - 2y. These intersect at
    z == -2 - 2y == 8 - x^2 - y^2 - 2y - 1
or
    x^2 + y^2 == 9
which is a circle of radius 3. Thus, the region R is defined in polar coordinates by
    0 <= r <= 3
    0 <= t <= 2Pi

volume = Integrate[1*r,{t,0,2Pi},{r,0,3},
      {z,-2-2r Sin[t],8-r^2 Cos[t]^2-(r Sin[t]+1)^2}]

N[%]

Here's a picture (not required):

ParametricPlot3D[
{r Cos[t], r Sin[t], 8-r^2 Cos[t]^2-(r Sin[t]+1)^2},
{r,0,3},{t,0,2Pi},
ViewPoint->{2.816, 1.415, 1.231}]

Problem 5

Find the volume of the solid enclosed by the spherical-coordinate surface
rho == 1 + Cos[phi].

Solution

In Assignment 9, Problem 5, you are asked to plot this surface in spherical coordinates.
For now we need only observe that the solid is described by the inequalities
    0 <=  rho  <= 1 + Cos[phi]
    0 <=  phi  <= Pi
    0 <= theta <= 2Pi
The volume element in spherical coordinates is rho^2 Sin[phi].

volume = Integrate[ rho^2 Sin[phi],
{theta,0,2Pi}, {phi,0,Pi}, {rho,0,1 + Cos[phi]}]