Find area of regions bounded by curve and line

DryRun
Gold Member
Messages
837
Reaction score
4
Homework Statement
Find area of regions bounded by
x^2 + y^2 = 9, y = 2x, x-axis in the first quadrant

The attempt at a solution
So, i drew the graph of y against x in my copybook, and circle with origin (0,0), radius = 3 units. The line y = 2x cuts through the circle.
Transforming to polar coordinates, the new limits are:
0≤θ≤tan^-1(3) and 0≤r≤3
\int\int rdrd\theta
After integration, i get the final answer: (9/2)tan^-1(3)
However, the answer in my notes is: (9/2)tan^-1(2). Did i copy the wrong answer or is my work wrong? I'm not sure.
 
Physics news on Phys.org
sharks said:
Homework Statement
Find area of regions bounded by
x^2 + y^2 = 9, y = 2x, x-axis in the first quadrant

The attempt at a solution
So, i drew the graph of y against x in my copybook, and circle with origin (0,0), radius = 3 units. The line y = 2x cuts through the circle.
Transforming to polar coordinates, the new limits are:
0≤θ≤tan^-1(3) and 0≤r≤3
\int\int rdrd\theta
After integration, i get the final answer: (9/2)tan^-1(3)
However, the answer in my notes is: (9/2)tan^-1(2). Did i copy the wrong answer or is my work wrong? I'm not sure.

Your answer is wrong, because the following is wrong
...the new limits are:
0≤θ≤tan^-1(3)...​
 
OK, so here is the graph:
http://s2.ipicture.ru/uploads/20111227/zpUVSESp.jpg
From my understanding, i need to find the area of the red section from the graph above.

For θ fixed, 0≤θ≤cos^-1(1/√5) and 0≤r≤3
I checked from calculator and cos^-1(1/√5) = tan^-1(2)

Thanks for your help, SammyS.:smile:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top