Calculate the following integral: double integral over D

Calculate the following integral: $$ \iint_D \frac{x}{y}\,dx\,dy, $$ where the region $D$ is bounded by the parabolas $y=x^2$ and $x=y^2$.

Step 1 · Find intersection points

Setting $y=x^2$ and $x=y^2$: substituting gives $x=(x^2)^2=x^4\Rightarrow x^4-x=0\Rightarrow x(x^3-1)=0\Rightarrow x=0$ or $x=1$. Intersection points: $(0,0)$ and $(1,1)$.

Step 2 · Determine which parabola is above

At $x=0.5$: $y=x^2=0.25$, while $y=\sqrt x\approx0.707$. Since $\sqrt x>x^2$ on $(0,1)$, the curve $y=\sqrt x$ (i.e. $x=y^2$) is on top.

Step 3 · Set up the integral

$$ I = \int_0^1\int_{x^2}^{\sqrt x} \frac{x}{y}\,dy\,dx. $$

Step 4 · Integrate with respect to $y$

$$ \int_{x^2}^{\sqrt x} \frac xy\,dy = x\left[\ln y\right]_{x^2}^{\sqrt x} = x\left(\ln\sqrt x - \ln x^2\right) = x\left(\frac12\ln x - 2\ln x\right) = -\frac{3x\ln x}{2}. $$

Step 5 · Integrate with respect to $x$

$$ I = -\frac32\int_0^1 x\ln x\,dx. $$

By parts with $u=\ln x$, $dv=x\,dx$ ($du=\frac{dx}x$, $v=\frac{x^2}{2}$):

$$ \int x\ln x\,dx = \frac{x^2\ln x}{2}-\int\frac{x}{2}\,dx = \frac{x^2\ln x}{2}-\frac{x^2}{4}. $$

At $x=1$: $0-\frac14=-\frac14$. At $x=0$: $\lim_{x\to0^+}\left(\frac{x^2\ln x}{2}-\frac{x^2}{4}\right)=0$ (since $x^2\ln x\to0$). So $\int_0^1 x\ln x\,dx=-\frac14$.

Step 6 · Final calculation

$$ I = -\frac32\cdot\left(-\frac14\right) = \frac38. $$

Additional Notes

• $\lim_{x\to0^+} x^2\ln x=0$ is a standard limit (polynomial growth dominates logarithmic decay), needed to evaluate the antiderivative at $x=0$.