# Triple integral calculation

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

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?

• the last limit should have only numbers at both ends, otherwise the volume would be a variable – vidyarthi Sep 24 at 6:52

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