${\int }_{-1}^1{\int }_{-\sqrt{1-x^2}\/2}^{\sqrt{1-x^2}\/2}F \mathrm{dy} \mathrm{dx}$ =  $\frac{\mathrm{\pi }}{3} \left(2\sqrt{2}-1\right)$ 

Tools≻Tasks≻Browse:

Calculus - Vector≻Integration≻Multiple Integration≻2-D≻Over an Ellipse

Table 6.2.8(a)   Task template for integration over an ellipse

$r$

$\=\frac{1}{2}\sqrt{\frac{{\tan }^2\left(\mathrm{\theta }\right)\+1}{\frac{1}{4}\+{\tan }^2\left(\mathrm{\theta }\right)}}$

$\=\sqrt{\frac{{\sec }^2\left(\mathrm{\θ}\right)}{1\+4 {\tan }^2\left(\mathrm{\θ}\right)}}$$$

$\=\sqrt{\frac{1}{{\cos }^2\left(\mathrm{\θ}\right)\+4 {\sin }^2\left(\mathrm{\θ}\right)}}$

$\=\frac{1}{\sqrt{1-{\sin }^2\left(\mathrm{\theta }\right)\+4 {\sin }^2\left(\mathrm{\theta }\right)}}$

$\=\frac{1}{\sqrt{1\+3 {\sin }^2\left(\mathrm{\θ}\right)}}$

$\genfrac{}{}{0ex}{}{x^2\+4 y^2\=1}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=r \cos \left(\mathrm{\theta }\right)\,y\=r \sin \left(\mathrm{\theta }\right)}$

$r^2{\cos \left(\mathrm{\θ}\right)}^2\+4r^2{\sin \left(\mathrm{\θ}\right)}^2\=1$

$\stackrel{\text{solutions for r}}{\to }$

$\frac{1}{\sqrt{1\+3{\sin \left(\mathrm{\θ}\right)}^2}}\,-\frac{1}{\sqrt{1\+3{\sin \left(\mathrm{\θ}\right)}^2}}$

Initialize

Loading Student:-MultivariateCalculus

$F\=\sqrt{2-x^2-4 y^2}$$\stackrel{\text{assign}}{\to }$

Solve for $y\=y\left(x\right)$ 

$x^2\+4 y^2\=1$$\stackrel{\text{solutions for y}}{\to }$$\frac{1}{2}\sqrt{-x^2\+1}\,-\frac{1}{2}\sqrt{-x^2\+1}$$\stackrel{\text{assign to a name}}{\to }$$Y$

Access the MultiInt command via the Context Panel

$F$ = $\sqrt{-x^2-4y^2\+2}$$\stackrel{\text{MultiInt}}{\to }$${\∫}_{-1}^1{\∫}_{-\frac{1}{2}\sqrt{-x^2\+1}}^{\frac{1}{2}\sqrt{-x^2\+1}}\sqrt{-x^2-4y^2\+2}\ⅆy\ⅆx$$\=$$\frac{2}{3}\mathrm{\π}\sqrt{2}-\frac{1}{3}\mathrm{\π}$$$

Write an appropriate iterated integral and evaluate

${\int }_{-1}^1{\int }_{Y_2}^{Y_1}F \mathit{\ⅆ}y \mathit{\ⅆ}x$$\stackrel{\text{simplify}}{\=}$$\frac{2}{3}\mathrm{\π}\sqrt{2}-\frac{1}{3}\mathrm{\π}$ $$

Table 6.2.8(b)   Solution from first principles

Tools≻Task≻Browse:

Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻Cartesian 2-D

Table 6.2.8(c)   Solution by visualization task template

Initialize

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{MultivariateCalculus}\right)\:$

$F≔\sqrt{2-x^2-4 y^2}\:$

Use the MultiInt command with a pre-defined domain option

$\mathrm{MultiInt}\left(F\,\left[x\,y\right]\=\mathrm{Ellipse}\left(x^2\+4 y^2\=1\right)\right)$ = $\frac{2}{3}\mathrm{\π}\sqrt{2}-\frac{1}{3}\mathrm{\π}$$$

Alternate solution(s)

$Y≔\mathrm{solve}\left(x^2\+4 y^2\=1\,y\right)$

$\frac{1}{2}\sqrt{-x^2\+1}\,-\frac{1}{2}\sqrt{-x^2\+1}$

Top-level, using the Int and int commands

$\mathrm{Int}\left(F\,\left[y\=Y_2..Y_1\,x\=-1..1\right]\right)\=\mathrm{int}\left(F\,\left[y\=Y_2..Y_1\,x\=-1..1\right]\right)\mathit{ }$

${\∫}_{-1}^1{\∫}_{-\frac{1}{2}\sqrt{-x^2\+1}}^{\frac{1}{2}\sqrt{-x^2\+1}}\sqrt{-x^2-4y^2\+2}\ⅆy\ⅆx\=\frac{2}{3}\sqrt{2}\mathrm{\π}-\frac{1}{3}\mathrm{\π}$

Iterate in the opposite order using top-level int and the evalf commands

$q≔\mathrm{int}\left(F\,\left[x\=-\sqrt{1-4 y^2}..\sqrt{1-4 y^2}\,y\=-1\/2..1\/2\right]\right)$

${\∫}_{-\frac{1}{2}}^{\frac{1}{2}}\left(-4\arcsin \left(\frac{\sqrt{-4y^2\+1}}{\sqrt{-4y^2\+2}}\right)y^2\+2\arcsin \left(\frac{\sqrt{-4y^2\+1}}{\sqrt{-4y^2\+2}}\right)\+\sqrt{-4y^2\+1}\right)\ⅆy$

$\mathrm{evalf}\left(q\right)$

$1.914724408$