Citat:
Ursprungligen postat av
melyhna
Detta,
\iint\! \frac{1}{\bigl(x^4 + 2 x^3 y + 3 x^2 y^2 + 2 x y^3 + y^4)^p}\,\mathrm{d}x\mathrm{d}y
&=
\iint\! \frac{4^p}{\bigl(r^4 \bigl(2 + \sin(2\theta)\bigr)^2\bigr)^p}r \,\mathrm{d}r\mathrm{d}\theta
Sätt
\[
\left\{
\begin{aligned}
x&=r\cos(\theta),\\
y&=r\sin(\theta),
\end{aligned}
\right.
\]
samt
\[
f(x,y)=x^4+2x^3y+3x^2y^2+2xy^3+y^4.
\]
Vi har att
\begin{align*}
h(r,\theta)&
=f(r\cos(\theta),r\sin(\theta))
\\&
=r^4\cos^4(\theta)+2r^3\cos^3(\theta)\cdot r\sin(\theta)+3r^2\cos^2(\theta)\cdot r^2\sin^2(\theta)+2r\cos(\theta)\cdot r^3\sin^3(\theta)+r^4\sin^4(\theta)
\\&
=r^4\bigl(\cos^4(\theta)+2\cos^3(\theta)\sin( \theta)+3\cos^2(\theta)\sin^2(\theta)+2\cos(\theta )\sin^3(\theta)+\sin^4(\theta)\bigr)
\\&
=r^4g(r,\theta)
\end{align*}
där
\begin{align*}
g(r,\theta)&
=
\cos^4(\theta)
+2\cos^3(\theta)\sin(\theta)
+3\cos^2(\theta)\sin^2(\theta)
+2\cos(\theta)\sin^3(\theta)
+\sin^4(\theta)
\\&=
\cos^2(\theta)\bigl(1-\sin^2(\theta)\bigr)
+\cos^2(\theta)\cdot2\sin(\theta)\cos(\theta)
+3\cos^2(\theta)\sin^2(\theta)
\\&\qquad{}
+\sin^2(\theta)\cdot2\sin(\theta)\cos(\theta)
+\sin^2(\theta)\bigl(1-\cos^2(\theta)\bigr)
\\&=
\cos^2(\theta)-\sin^2(\theta)\cos^2(\theta)
+\cos^2(\theta)\sin(2\theta)
+3\sin^2(\theta)\cos^2(\theta)
\\&\qquad{}
+\sin^2(\theta)\sin(2\theta)
+\sin^2(\theta)-\sin^2(\theta)\cos^2(\theta)
\\&=
\sin^2(\theta)+\cos^2(\theta)
+\sin^2(\theta)\cos^2(\theta)
+\bigl(\sin^2(\theta)+\cos^2(\theta)\bigr)\sin(2 \theta)
\\&=1+\sin^2(\theta)\cos^2(\theta)+\sin(2\theta)
\\&=\tfrac{1}{4}\bigl(4+4\sin^2(\theta)\cos^2( \theta)+4\sin(2\theta)\bigr)
\\&=\tfrac{1}{4}\bigl(2^2+\bigl(2\sin(\theta)\cos( \theta)\bigr)^2+2\cdot2\cdot\sin(2\theta)\bigr)
\\&=\tfrac{1}{4}\bigl(2^2+2\cdot2\cdot\sin(2\theta )+\bigl(2\sin(\theta)\cos(\theta)\bigr)^2\bigr)
\\&=\tfrac{1}{4}\bigl(2^2+2\cdot2\cdot\sin(2\theta )+\sin^2(2\theta)\bigr)
\\&=\tfrac{1}{4}\bigl(2+\sin(2\theta)\bigr)^2
\end{align*}
varför
\[
h(r,\theta)=\tfrac{1}{4}r^4\bigl(2+\sin(2\theta) \bigr)^2
\]
vilket ger
\begin{align*}
\iint\!\frac{1}{f(x,y)^p}\,\mathrm{d}x\mathrm{d}y
&=\iint\!\frac{1}{h(r,\theta)^p}\cdot r\,\mathrm{d}r\mathrm{d}\theta
=\iint\!\frac{1}{\bigl(\tfrac{1}{4}r^4\bigl(2+\sin (2\theta)\bigr)^2\bigr)^p}\cdot r\,\mathrm{d}r\mathrm{d}\theta
\\&=\iint\! \frac{1}{r^{4p-1}}\cdot\frac{4^p}{\bigl(2 + \sin(2\theta)\bigr)^{2p}} \,\mathrm{d}r\mathrm{d}\theta
=\int_0^1\!\frac{1}{r^{4p-1}}\,\mathrm{d}r
\int_0^{2\pi}\!\frac{4^p}{\bigl(2 + \sin(2\theta)\bigr)^{2p}}\,\mathrm{d}\theta
\\&=\int_0^1\!r^{1-4p}\,\mathrm{d}r\cdot A
=\Bigl[\frac{r^{2-4p}}{2-4p}\Bigr]_0^1\cdot A
\end{align*}
som är ändlig om
\[
2-4p>0
\quad\Leftrightarrow\quad
0<p<\tfrac{1}{2}.
\]
