Finding the Area of a Polar Function: Are Your Limits of Integration Correct?

[Bryon]

My question here is do I have the correct limits of integration? At first I thought it would be from pi/10 to 3pi/10 but I have a feeling that those are incorrect.

Homework Statement

Find the area of one petal of the polar function r(x) = cos(5x)

Homework Equations

integral[alpa to beta] .5* r(x)^2dx

The Attempt at a Solution

cos(5x) = 0 when x = (1/5)*pi/2 = pi/10
This means that the limits of integration are pi/10 and -pi/10

integral (.5*cos(5x))dx = 1/4x - (sin(5x)/20) from pi/10 to -pi/10

 

[Mark44]

My question here is do I have the correct limits of integration? At first I thought it would be from pi/10 to 3pi/10 but I have a feeling that those are incorrect.

Homework Statement

Find the area of one petal of the polar function r(x) = cos(5x)

Homework Equations

integral[alpa to beta] .5* r(x)^2dx

The Attempt at a Solution

cos(5x) = 0 when x = (1/5)*pi/2 = pi/10
This means that the limits of integration are pi/10 and -pi/10

Yes, these are correct.

integral (.5*cos(5x))dx = 1/4x - (sin(5x)/20) from pi/10 to -pi/10

You have it in your relevant equations, but you forgot to square r(x) in the integral just above. Or maybe you just forgot to put in the exponent in your integrand.

 

[Bryon]

Oops...sorry cos(5x)^2 = (1+cos(5x))/2

Thanks! for some reason I was thinking it was between pi/10 and 3pi/10.

 

[Mindscrape]

Thanks! for some reason I was thinking it was between pi/10 and 3pi/10.

That would work too. :)