Consider the volume, $V$, enclosed by the paraboloid $$z=x^2+y^2$$ and the plane $2x+z=3$. Determine the limits for the enclosed volume. Do not attempt to evaluate the integral.
Hello, I’m confused on how to get limits for this integral, I drew it and I think $z$ limits are from the paraboloid equation to $z=3-2x$, but after that I’m stumped.
Thanks!
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Well, first we can find the $x$ and $y$ coordinates that we'll have to integrate across by solving for the intersection of the plane and the paraboloid. To do this, we rearrange our equations as $z=3-2x$ and $z=x^2+y^2$ and set them equal to get $$3-2x=x^2+y^2 \implies 4=(x+1)^2+y^2$$ So our region is a circle of radius $2$ around the point $(x,y)=(-1,0)$. Now we have our limits by rearranging our circle into $x=-1 \pm\sqrt{4-y^2}$. The $z$ values that we will integrate across will simply be from the paraboloid to the plane so we can write $$\int_{-2}^2 \int_{-1-\sqrt{4-y^2}}^{-1+\sqrt{4-y^2}}\int_{x^2+y^2}^{3-2x} dz \ dx \ dy$$ or $$\int_{-2}^2 \int_{-1-\sqrt{4-y^2}}^{-1+\sqrt{4-y^2}}(3-2x) - (x^2+y^2) \ dx \ dy$$
The two surfaces intersect when $$z=x^2+y^2=3-2x \Rightarrow x^2+2x+y^2 = 3$$ Now $$x^2+2x+y^2 = 3 \Rightarrow \left(x+1\right)^2+y^2=4$$
Therefore in cartesian coordinates: $$V= \int{\int{\int dz}dy}dx=\int_{x=-3}^{x=1}{\int_{y=-\sqrt{4-(x+1)^2}}^{y=\sqrt{4-(x+1)^2}}{\int_{z=x^2+y^2}^{z=3-2x}1dz}dy}dx$$
In Cylindrical coordinates, starting from: $$V= \int{\int{\int_{z=x^2+y^2}^{z=3-2x}1dz}}dA=\int{\int{\left[4-\left(x+1\right)^2-y^2\right]}}dA$$ Then using $r=x^2+y^2$, $x=r\cos\theta$ and $y=r\sin\theta$: $$V=\int_{\theta=0}^{\theta=2\pi}{\int_{r=0}^{r=2}{\left[ {4-\left(r\cos\theta+1\right)^2-\left( {r\sin\theta} \right)^2} \right]}}rdrd\theta$$