# Find the volume of the solid bounded by the xy plane, the cylinder $x^{2} + y^{2}=4$, and the plane $z+y=4$.

Find the volume of the solid bounded by the xy plane, the cylinder $$x^{2} + y^{2}=4$$, and the plane $$z+y=4$$.

If we draw the graph, then the integral will be calculated should be

$$\int_{0}^{2} \int_{0}^{\sqrt{4-x^{2}}} (4-y) \: dy dx$$

This would give the volume of the solid in 1st quadrant which can also be obtained through

$$\int_{0}^{2} \int_{0}^{\sqrt{4-y^{2}}} (4-y) \: dx dy$$

which would be equal to $$4\pi - \frac83$$. Both the above equations give the same result.

But if we try to find the volume of the entire solid formed by the three curves, then the results from the above equations(after changing the limits appropriately) don't match. $$\int_{-2}^{2} 2\int_{0}^{\sqrt{4-x^{2}}} (4-y) \: dy dx = 2(8\pi-\frac{16}{3})$$ $$\int_{-2}^{2} 2\int_{0}^{\sqrt{4-y^{2}}} (4-y) \: dx dy = 16\pi$$

why the results of the these two equations don't match? Is there some problem with the limits I've set or something else? Please help

PS: There is a similar question, but my query is different.

The limits of the second integral are wrong. It should be $$2\int_0^2\int_{-\sqrt{4-y^{2}}}^{\sqrt{4-y^{2}}} (4-y) \: dx\ dy=4\int_0^2\int_0^{\sqrt{4-y^{2}}} (4-y) \: dx\ dy$$ The answers match.