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Question Number 73428 by Learner-123 last updated on 12/Nov/19

$$Evaluate:$$$$1){\int }_{-2}^2{\int }_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}(3-x)dydx.$$$$\left(afterchangingtheintegraltopolarform\right).$$$$$$$$2){\int }_0^4{\int }_0^{4-x}{\int }_0^{4-\frac{y^2}{4}}dzdydx.$$

Commented by mathmax by abdo last updated on 12/Nov/19

$$1)I={\int }_{-2}^2\left({\int }_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}dy\right)(3-x)dx={\int }_{-2}^{2}2\sqrt{4-x^2}(3-x)dx$$$$=2{\int }_{-2}^2(3-x)\sqrt{4-x^2}dx{=}_{x=2sin\theta }2{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(3-2sin\theta )2cos\theta \left(2cos\theta \right)d\theta$$$$=8{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(3-2sin\theta ){cos}^2\theta d\theta =24{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{cos}^2\theta d\theta -16{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}sin\theta {cos}^2\theta d\theta$$$$wehave{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}sin\theta {cos}^2\theta d\theta ={[-\frac{1}{3}{cos}^3\theta ]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}=0$$$${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{cos}^2\theta d\theta =2{\int }_0^{\frac{\pi }{2}}\frac{1+cos\left(2\theta \right)}{2}d\theta =\frac{\pi }{2}+{\int }_0^{\frac{\pi }{2}}cos\left(2\theta \right)d\theta$$$$=\frac{\pi }{2}+{\left[\frac{1}{2}sin\left(2\theta \right)\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{2}+0=\frac{\pi }{2}\Rightarrow I=24\times \frac{\pi }{2}\Rightarrow I=12\pi$$

Commented by Learner-123 last updated on 12/Nov/19

$$thnkssir.$$

Answered by Henri Boucatchou last updated on 12/Nov/19

$$1)I={\int }_{-2}^{2}2(3-x)\sqrt{4-x^2}dx$$$$x=2cos\theta ,-1⩽cos\theta ⩽1,dx=-2sin\theta d\theta$$$$I=4{\int }_{-\pi }^0(3-2cos\theta )\mid sin\theta \mid (-2sin\theta )d\theta$$$$=8\left[{\int }_{-\pi }^0(3-2cos\theta ){sin}^2\theta d\theta \right]$$$$=8[3{\int }_{-\pi }^0{sin}^2\theta d\theta -2{\int }_{-\pi }^{0}cos\theta {sin}^2\theta d\theta ]$$$$=8[\frac{3}{2}{\int }_{-\pi }^0(1-cos2\theta )d\theta -2.0]$$$$=8.\frac{3}{2}{[\theta -\frac{1}{2}sin2\theta ]}_{-\pi }^0$$$$=12\pi$$$$---------------$$$${\int }_0^4{\int }_0^{4-x}{\int }_0^{4-y^2/4}dzdydx=$$$${\int }_0^4{\int }_0^{4-x}(4-\frac{y^2}{4})dydx=$$$${\int }_0^4{[4y-\frac{y^3}{12}]}_0^{4-x}dx=$$$${\int }_0^4(4(4-x)-\frac{1}{12}{(4-x)}^3)dx=$$$${4(x-\frac{1}{2}{x}^2)+\frac{1}{12.4}{(4-x)}^4]}_0^4=$$$$4(4-\frac{1}{2}{4}^2)-\frac{1}{12}=$$$$-16-\frac{1}{12}=$$$$-\frac{193}{12}$$$$$$$$\mathit{P}\mathit{L}\mathit{E}\mathit{A}\mathit{S}\mathit{E}\mathit{C}\mathit{H}\mathit{E}\mathit{C}\mathit{K}\mathit{I}\mathit{F}{\mathit{T}\mathit{H}\mathit{E}\mathit{R}\mathit{E}}^{\prime }\mathit{S}\mathit{N}\mathit{O}\mathit{E}\mathit{R}\mathit{R}\mathit{O}\mathit{R}$$$$$$

Commented by Learner-123 last updated on 12/Nov/19

$${ma}^{\prime }am,thereisslightmistakein4thlast$$$$line.$$

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