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Iterated Integral

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${\int }_0^1{\int }_{{\sqrt{1-x^2}}_{}}^1\left(1\+2 x^2\+3 y^2\right) \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{8}{3}-\frac{9}{16}\mathrm{\π}$$$$$

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${\int }_0^1{\int }_{-1}^{-\sqrt{1-x^2}}\left(1\+2 x^2\+3 y^2\right) \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{8}{3}-\frac{9}{16}\mathrm{\π}$$$

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${\int }_3^4{\int }_{{\sqrt{1-{\left(x-4\right)}^2}}_{}}^1\left(1\+2 x^2\+3 y^2\right) \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $32-\frac{137}{16}\mathrm{\π}$$$

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${\int }_3^4{\int }_{-1}^{-\sqrt{1-{\left(x-4\right)}^2}}\left(1\+2 x^2\+3 y^2\right) \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $32-\frac{137}{16}\mathrm{\π}$$$

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${\int }_1^3{\int }_{-1}^1\left(1\+2 x^2\+3 y^2\right) \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{128}{3}$$$

Table 5.3.5(a)   Five iterated integrals for the integration order $\mathrm{dy} \mathrm{dx}$ 

Initialize

Loading Student:-MultivariateCalculus

$f\=1\+2 x^2\+3 y^2$$\stackrel{\text{assign}}{\to }$

$\mathrm{XL}\=\sqrt{1-y^2}$$\stackrel{\text{assign}}{\to }$

$\mathrm{XR}\=4-\sqrt{1-y^2}$$\stackrel{\text{assign}}{\to }$

Access the MultiInt command via the Context Panel

$f$ = $2x^2\+3y^2\+1$$\stackrel{\text{MultiInt}}{\to }$${\∫}_{-1}^1{\∫}_{\sqrt{-y^2\+1}}^{4-\sqrt{-y^2\+1}}\left(2x^2\+3y^2\+1\right)\ⅆx\ⅆy$$\=$$112-\frac{73}{4}\mathrm{\π}$

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Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻

Table 5.3.5(b)   Visualizing $R$ and the resulting volume for iteration in the order $\mathrm{dy} \mathrm{dx}$ 

Iterate in the order $\mathrm{dy} \mathrm{dx}$ 

${\int }_{-1}^1{\int }_{{\sqrt{1-y^2}}_{}}^{4-\sqrt{1-y^2}}f \mathit{\ⅆ}x \mathit{\ⅆ}y$ = $112-\frac{73}{4}\mathrm{\π}$$$

Display the iterated integrals

${\int }_{-1}^1{\int }_{{\sqrt{1-y^2}}_{}}^{4-\sqrt{1-y^2}}f \mathit{\ⅆ}x \mathit{\ⅆ}y$

${\∫}_{-1}^1{\∫}_{\sqrt{-y^2\+1}}^{4-\sqrt{-y^2\+1}}\left(2x^2\+3y^2\+1\right)\ⅆx\ⅆy$

$\=$

$112-\frac{73}{4}\mathrm{\π}$

Initialize

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

$f≔1\+2 x^2\+3 y^2\:$

Iterate in the order $\mathrm{dx} \mathrm{dy}$

$\mathrm{Int}\left(f\,\left[x\=\sqrt{1-y^2}..4-\sqrt{1-y^2}\,y\=-1..1\right]\right)\=\mathrm{int}\left(f\,\left[x\=\sqrt{1-y^2}..4-\sqrt{1-y^2}\,y\=-1..1\right]\right)$

${\∫}_{-1}^1{\∫}_{\sqrt{-y^2\+1}}^{4-\sqrt{-y^2\+1}}\left(2x^2\+3y^2\+1\right)\ⅆx\ⅆy\=112-\frac{73}{4}\mathrm{\π}$

Use the MultiInt command from the Student MultivariateCalculus package

$\mathrm{MultiInt}\left(f\,x\=\sqrt{1-y^2}..4-\sqrt{1-y^2}\,y\=-1..1\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_{-1}^1{\∫}_{\sqrt{-y^2\+1}}^{4-\sqrt{-y^2\+1}}\left(2x^2\+3y^2\+1\right)\ⅆx\ⅆy$

$\mathrm{MultiInt}\left(f\,x\=\sqrt{1-y^2}..4-\sqrt{1-y^2}\,y\=-1..1\right)$

$112-\frac{73}{4}\mathrm{\π}$

Obtain stepwise evaluations via the MultiInt command

$\mathrm{MultiInt}\left(f\,x\=\sqrt{1-y^2}..4-\sqrt{1-y^2}\,y\=-1..1\,\mathrm{output}\=\mathrm{steps}\right)$:

${\int }_{-1}^1{\int }_{\sqrt{-y^2\+1}}^{4-\sqrt{-y^2\+1}}\left(2x^2\+3y^2\+1\right)\ⅆx\ⅆy$

$\=112-\frac{73}{4}\mathrm{\π}$