Solving for Area of a Given Curve: sqrt(|x|) + sqrt(|y|) = 1

[soofjan]

Homework Statement

Find the area, whose edge is given by the following curve:
sqrt(|x|) + sqrt(|y|) = 1
Also, draw the area.
Guidance: try x = r*cos(t)^k, y = r*sin(t)^k for a fitting k.

Homework Equations

The Attempt at a Solution

I tried:
x = r*cos(t)^4, y = r*sin(t)^4
From the curve equation, I get that: sqrt(r)=1. So:
0 \leq r \leq 1
0 \leq t \leq 2*pi
|J| = 4r*|sin(t)^3*cos(t)^3|.
Since I have an absolute value as the integrand, I calculate the integral of t from 0 to pi/2, and multiply the result by 4. The final answer is 2/3.

I believe that my transformation is wrong, because when I try to draw it via Wolfram, it gives me a circle, but it is supposed to look like an astroid.

Any help would be appreciated. Thanks!

 

[Dick]

Homework Statement

Find the area, whose edge is given by the following curve:
sqrt(|x|) + sqrt(|y|) = 1
Also, draw the area.
Guidance: try x = r*cos(t)^k, y = r*sin(t)^k for a fitting k.

Homework Equations

The Attempt at a Solution

I tried:
x = r*cos(t)^4, y = r*sin(t)^4
From the curve equation, I get that: sqrt(r)=1. So:
0 \leq r \leq 1
0 \leq t \leq 2*pi
|J| = 4r*|sin(t)^3*cos(t)^3|.
Since I have an absolute value as the integrand, I calculate the integral of t from 0 to pi/2, and multiply the result by 4. The final answer is 2/3.

I believe that my transformation is wrong, because when I try to draw it via Wolfram, it gives me a circle, but it is supposed to look like an astroid.

Any help would be appreciated. Thanks!

That all looks ok to me. Except that my plot doesn't look like a circle. What are you plotting?

 

[soofjan]

I tried to polar plot the following on Wolfram:
sqrt(|r*cos(t)^4|) + sqrt(|r*sin(t)^4|) = 1
r=0..1, t=0..2*pi

 

[Dick]

I tried to polar plot the following on Wolfram:
sqrt(|r*cos(t)^4|) + sqrt(|r*sin(t)^4|) = 1
r=0..1, t=0..2*pi

I'm not sure what that would mean to Wolfram. What you want to plot is the set of all (x,y) SUCH THAT sqrt(|x|)+sqrt(|y|)=1. I.e. parametric plot x(t)=cos(t)^4, y(t)=sin(t)^4, for t in [0,pi/2], the actual x,y values, not their square roots.

 

[soofjan]

I see your point. Thanks again.