<s>[INST] How to find limits of integral to find volume? Find the volume generated by the plane region, in the first quadrant, bounded by the graph of the function $ y=\sqrt{9-x^2} $ sbout the y-axis.
I know how to solve it using the formula but how do I get the upper and lower limits of the integration.
 [/INST] If I understood the question correctly, the volume of the solid they're asking you to find is a hemisphere. If you have to rotate it about the y-axis, use the shell method:
$$
V=2\pi\int_{0}^{3}x\sqrt{9-x^2}\,dx
$$
Your bounds of integration are from $0$ to $3$ because $\sqrt{9-x^2}$ is a circle of radius $3$ centered at the origin.
The integral itself:
$$
\int x\sqrt{9-x^2}\,dx=
-\frac{1}{2}\int\sqrt{9-x^2}\frac{d}{dx}(9-x^2)\,dx=\\
-\frac{1}{2}\int\sqrt{9-x^2}\,d(9-x^2)=
-\frac{1}{2}\frac{2\sqrt{(9-x^2)^3}}{3}=\\
-\frac{\sqrt{(9-x^2)^3}}{3}+C.
$$
The volume:
$$
V=2\pi\int_{0}^{3}x\sqrt{9-x^2}\,dx=
-2\pi\frac{\sqrt{(9-x^2)^3}}{3}\bigg|_{0}^{3}=\\
-2\pi\left(\frac{\sqrt{(9-3^2)^3}}{3}-\frac{\sqrt{(9-0^2)^3}}{3}\right)=\\
-2\pi\left(0-\frac{9\cdot3}{3}\right)=-2\pi(-9)=18\pi\ cubic\ units.
$$
</s>