What's wrong with the calculation with polar coordinates here?

Question

Suppose $x=r\cos t$ and $y=r\sin t$. I did the following calculation:

$$ \begin{align} &x^4+y^4=(r\cos t)^4+(r\sin t)^4\\ =&r^4(\sin^4t+\cos^4t)\\ =&r^4[(\sin^2t+\cos^2t)^2-2\sin^2t\cos^2t]\\ =&r^4(1-2\sin^2t\cos^2t)=r^4-2r^4\sin^2t\cos^2t \end{align} $$ On the other hand $$ \begin{align} x^4+y^4&=(x^2+y^2)^2-2x^2y^2\\ &=r^4-2r^2\sin^2t\cos^2t \end{align} $$

What is wrong with the calculation?

Note $x^{2}y^{2} = (r\cos t)^{2}(r\sin t)^{2} = (r^{2}\cos^{2} t)(r^{2}\sin^{2} t) = r^{4}\cos^{2} t\sin^{2}t$. — Alex Wertheim, Nov 05 '14 at 20:25

score 2 · Accepted Answer · answered Nov 05 '14 at 20:26

2

Notice that $2x^2y^2 = 2(r\sin t)^2(r\cos t)^2 = 2r^4\sin^2t\cos^2t$.

answered Nov 05 '14 at 20:26

apnorton

17,706

What's wrong with the calculation with polar coordinates here?

1 Answers1