Essentially you want to integrate $\int_0^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} z dy dx$ where $z = \sqrt{9 - y^2 - x^2}$. What geometric object is this? It might be of help to rewrite it as $x^2 + y^2 + z^2 = 9$
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The equation $x^2 + y^2 + z^2 = 9$ represents a sphere in $3$D. The limits of the integral will give you the portion of the sphere over which you are integrating and hence the integral will give you the volume of the portion of the sphere. In this case, $z$ is always positive. This restricts the object to a hemi-sphere. The $y$ takes negative and positive values. The $x$ takes only positive values which again restricts the object to half of the hemi-sphere. Hence, the portion of the sphere you are integrating is just $1/4$ of the total sphere. The total volume of the sphere is $\dfrac43 \times \pi \times 3^3$. Hence, the integral is $\dfrac{\dfrac43 \times \pi \times 3^3}{4} = 9 \pi$