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$$ \text { Find the volume of the solid bounded by the surfaces } x=0 \text { and } y^{2}+z^{2}=4 \text { and } x+z=4 $$

The part which I am confused is what are limits of integration over here $x$ goes from $ 0 $ to $4-z$ ; $ y $ goes from $0$ to $ 4 - z^2$ and $ z $ from $0$ to $z$. Is it correct?

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  • $\begingroup$ the last limit should have only numbers at both ends, otherwise the volume would be a variable $\endgroup$ – vidyarthi Sep 24 at 6:52
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We can refer to the $y, z$ plane and the integration boundary is a circle centered in the origin with radius $r=2$.

Then x varies from $0$ to $4-z$.

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    $\begingroup$ @mathsstudent It is probably $\int_0^2\int_0^{\sqrt{4-z^2}}\int_0^{4-z}dxdydz$ $\endgroup$ – vidyarthi Sep 24 at 6:56
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    $\begingroup$ @vidyarthi In the polar form the term “4-r” should be $4-r\sin \theta$. $\endgroup$ – user Sep 24 at 7:12
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    $\begingroup$ @mathsstudent Sorry I see it now but the correct set up in cartesian should be $$\int_{-2}^2\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}\int_0^{4-y}dxdydz$$ $\endgroup$ – user Sep 24 at 7:38
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    $\begingroup$ @mathsstudent I confirm also that the polar form $$\int_0^2\int_0^{2\pi}\int_0^{4-r \sin \theta}rdxd\theta dr$$ is correct. You should obtain the same result. $\endgroup$ – user Sep 24 at 7:41
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    $\begingroup$ @mathsstudent Anyway by symmetry we can argue that the volume is an half cylinder with area $4\pi$ and hight 8. $\endgroup$ – user Sep 24 at 7:42

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