# Multi variable integral : $\int_0^1 \int_\sqrt{y}^1 \sqrt{x^3+1} \, dx \, dy$

$$\int_0^1 \int_\sqrt{y}^1 \sqrt{x^3+1} \, dx \, dy$$ Here is my problem in my workbook. If I solve this problem by definition, that find integral for $x$, after that solve for $y$. so $\int_\sqrt{y}^1 \sqrt{x^3+1} \, dx$ is so complicate. I have used Maple, but the result still long and complicate that I cannot use it to find integral for $y$.

Thanks :)

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Switch the order of integration. (Do $y$ first, then $x$.) – Hans Lundmark Jun 12 '12 at 19:10

The way you have the integration setup, you integrate along $x$ first for a fixed $y$. The figure below indicates how you would go about with the integration. You integrate over the horizontal red strip first and then move the horizontal strip from $y=0$ to $y=1$. Now for the ease of integration, change the order of integration and integrate along $y$ first for a fixed $x$. The figure indicates how you would go about with the integration. You integrate over the vertical red strip first and then move the vertical strip from $x=0$ to $x=1$. Hence, if you swap the integrals the limits become $y$ going from $0$ to $x^2$ and $x$ goes from $0$ to $1$. $$I = \int_0^1 \int_\sqrt{y}^1 \sqrt{x^3+1} dx dy = \int_0^1 \int_0^{x^2} \sqrt{x^3+1} dy dx = \int_0^1 x^2 \sqrt{x^3+1} dx$$ Now call $x^3+1 = t^2$. Then we have that $3x^2 dx = 2t dt \implies x^2 dx = \dfrac{2}{3}tdt$. As $x$ varies from $0$ to $1$, we have that $t$ varies from $1$ to $\sqrt{2}$. Hence, we get that $$I = \int_1^\sqrt{2}\dfrac23 t \times t dt = \dfrac23 \int_1^\sqrt{2}t^2 dt = \dfrac23 \times \left. \dfrac{t^3}3\right \vert_{t=1}^{t=\sqrt{2}} = \dfrac29 \left( (\sqrt{2})^3 - 1^3\right) = \dfrac29 (2 \sqrt{2} - 1).$$
@Marvis region over which you are integrating is below the parabola $y=x^2$ from x=0 to 1 Could you explain more, please. – hqt Jun 13 '12 at 3:01