${\int }_{-1}^1{\int }_{-\sqrt{1-x^2}\/2}^{\sqrt{1-x^2}\/2}1 \mathrm{dy} \mathrm{dx}$ =  $\frac{\mathrm{\π}}{2}$ 

Initialize

Tools≻Load Package: Student Multivariate Calculus

Loading Student:-MultivariateCalculus

Access the MultiInt command via the Context Panel

$c\=\sqrt{1-x^2}\/2$$\stackrel{\text{assign}}{\to }$

$1$$\stackrel{\text{MultiInt}}{\to }$${\int }_{\mathrm{-1}}^1{\int }_{-\frac{\sqrt{-x^2+1}}{2}}^{\frac{\sqrt{-x^2+1}}{2}}1\ⅆy\ⅆx$$\=$$\frac{\mathrm{\pi }}{2}$

Tools≻Tasks≻Browse:

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

Table 6.1.8(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)}}$$$

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

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

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

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

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

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

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

Solve for $y\=y\left(x\right)$ 

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

Write an appropriate iterated integral and evaluate

${\int }_{-1}^1{\int }_{Y_2}^{Y_1}1 \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{\mathrm{\pi }}{2}$$$

Table 6.1.8(b)   Solution from first principles

Tools≻Task≻Browse:

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

Table 6.1.8(c)   Solution by visualization task template

Initialize

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

Solve for $y\=y\left(x\right)$ 

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

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

Top-level, using the Int and int commands

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

${\int }_{\mathrm{-1}}^1{\int }_{-\frac{\sqrt{-x^2+1}}{2}}^{\frac{\sqrt{-x^2+1}}{2}}1\ⅆy\ⅆx=\frac{\mathrm{\pi }}{2}$

Use the MultiInt command from the Student MultivariateCalculus package

$\mathrm{MultiInt}\left(1\,\left[x\,y\right]\=\mathrm{Ellipse}\left(x^2\+4 y^2\=1\right)\right)$ = $\frac{\mathrm{\pi }}{2}$$$

$\mathrm{MultiInt}\left(1\,\left[x\,y\right]\=\mathrm{Ellipse}\left(x^2\+4 y^2\=1\,\left[r\,\mathrm{\θ}\right]\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\int }_0^{2\mathrm{\pi }}{\int }_0^{\frac{\sqrt{\frac{{\tan \left(\mathrm{\theta }\right)}^2+1}{\frac{1}{4}+{\tan \left(\mathrm{\theta }\right)}^2}}}{2}}r\ⅆr\ⅆ\mathrm{\theta }$

$\mathrm{MultiInt}\left(1\,\left[x\,y\right]\=\mathrm{Ellipse}\left(x^2\+4 y^2\=1\,\left[r\,\mathrm{\theta }\right]\right)\,\mathrm{output}\=\mathrm{steps}\right)$

$\frac{\mathrm{\pi }}{2}$