Multiple integral : $\iint_{D}\sqrt{1-x^2-y^2}dxdy=\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\Big(\int_{-1}^{1}\sqrt{1-x^2-y^2}dx\Big)dy$

Question

I would like to compute the volume of $$ E=\{(x,y,z) \in \mathbb{R}^3 \mid (x,y) \in D \text{ and }0\leq z \leq \sqrt{1-x^2-y^2}\} $$ where the region $D=\{(x,y) \in \mathbb{R}^2 \mid x^2+y^2 \leq 1\}$.

I know that I have to compute the following integral : $$\iint_{D}\sqrt{1-x^2-y^2}dxdy=\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\Big(\int_{-1}^{1}\sqrt{1-x^2-y^2}dx\Big)dy$$ And $$\int_{-1}^{1}\sqrt{1-x^2-y^2}dx=\frac{x}{2}\sqrt{1-y^2-x^2}+\frac{1-y^2}{2}\arcsin\frac{x}{\sqrt{1-y^2}}\Big|_{-1}^{1}$$ But this leads to $\sqrt{-y^2}$... I don't see where is my mistake ? Because when I compute : $$\int_{-1}^{1}\Big(\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\sqrt{1-x^2-y^2}dx\Big)dy$$ I don't have any problem and i find $\frac{2\pi}{3}$ which is correct. Thank you for your help !

Answer 1

You have $$ D= \left\{ (x,y) \in\mathbb R^2: -1 \le x\le 1, -\sqrt{1-x^2} \le y \le \sqrt{1-x^2}\right\}$$ Pleas note that the limits on $y$ depends on $x$; that means that you cannot integrate over $x$ before you integrate over $y$ - the integration over $y$ has to be performed first. The correct way to write your integral is: $$ \iint_D \sqrt{1-x^2-y^2} dxdy = \int_{-1}^1 \left(\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \sqrt{1-x^2-y^2} dy\right) dx$$ If you really want to make the integration over $x$ first, you need to represent the integration area in a different way: $$ D= \left\{ (x,y) \in\mathbb R^2: -1 \le y\le 1, -\sqrt{1-y^2} \le x \le \sqrt{1-y^2}\right\}$$ which gives you the other representation of the integral, which, as you noticed, gives the correct result: $$ \iint_D \sqrt{1-x^2-y^2} dxdy = \int_{-1}^1 \left(\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \sqrt{1-x^2-y^2} dx\right) dy$$

Answer 2

Actually, the integral that you should compute is$$\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2-y^2}}1\,\mathrm dz\,\mathrm dy\,\mathrm dx=\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\sqrt{1-x^2-y^2}\,\mathrm dy\,\mathrm dx.$$But it is much easier to do it in spherical coordinates:$$\int_0^{2\pi}\int_0^{\pi/2}\int_0^1\rho^2\sin(\varphi)\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta.$$You will get $\frac23\pi$.

Answer 3

Note that the correct usage of Fubini would be: $$ \begin{aligned} \iint_{D}\sqrt{1-x^2-y^2}\; dx\; dy &= \int_{-1}^{1}dx \int_{-\sqrt{1-x^2}}^{+\sqrt{1-x^2}}dy\; \sqrt{1-x^2-y^2} \\ &\qquad\text{since $y$ between $\pm \sqrt{1-x^2}$} \\ &= \int_{-1}^{1}dx\; \frac \pi 2(1-x^2) \\ &=\frac{2\pi}3\ , \end{aligned} $$ corresponding to the half volume of the $3$-ball.