Polar coordinates finding area between two curves

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Homework Help Overview

The discussion revolves around finding the area between two curves represented in polar coordinates, specifically the equations r = sin(x) and r = cos(x). Participants are exploring how to set up the integral for this area calculation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants suggest drawing the curves to better understand the integral setup. There are attempts to determine the intersection points of the curves and discussions about converting to rectangular coordinates. Questions arise regarding the appropriate interval for integration and the implications of using the full range of 0 to 2π.

Discussion Status

The discussion is active, with various approaches being considered. Some participants have provided guidance on the importance of understanding the interval and the behavior of the curves, while others are questioning the assumptions about the integration limits.

Contextual Notes

There is a mention of the curves intersecting at specific angles and the need to consider the behavior of sine and cosine over the chosen interval. The problem is identified as part of a Calculus 2 course.

th3plan
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Homework Statement





Homework Equations



r=sinx
r= cosx

Ok , i need help how to properly select the integral to evaluate the area they make. Can someone please show me how , i know how to evaluate it just having hard times with integrals

The Attempt at a Solution

 
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it will be easiest if you first draw out the curves. This will help you figure out what your integral should be.
 
i graphed it on calculator, and i did set them equal to each other to get tanx=1 but from here on idk what to do
 
first of all over what interval?
 
if you're given those in parametric form in polar, you are going to get two circles in the x,y (or r,theta) plane, i believe. But as ice109 said, its also important that you know how x varies for this one.
 
the interval is 0,2pi
 
Multiply both sides by r, then change to rectangular form. Is this Calculus 2 or 3? B/c I did this problem yesterday.
 
Calc 2
 
how would i change it to rectangular form, using x=rcos(theta) and y=rsin(theta)

?
 
  • #10
th3plan said:
the interval is 0,2pi
No, it isn't. Since sine and cosine are negative for half that interval using 0 to 2[itex]\pi[/itex] gives you each circle twice. And, in fact, the area you want only requires [itex]\theta[/itex] going from 0 to [itex]\pi/2[/itex].

However, you are correct that the circles intersect when tan[itex]\theta[/itex]= 1- that is, at [itex]\theta= \pi/4[/itex] as well as at 0. For [itex]0\le \theta\le \pi/4[/itex], a radius goes from 0 to cos([itex]\theta[/itex]) while from [itex]\pi/4\le \theta\le \pi/2[/itex] it goes from 0 to sin([itex]\theta[/itex]). From symmetry, you should be able to integrate cos([itex]\theta[/itex]) from 0 to [itex]\pi/4[/itex] and double.
 

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