Question

(a) Find the volume of the region \({\rm{E}}\)bounded by the paraboloids \({\rm{z = }}{{\rm{x}}^{\rm{2}}}{\rm{ + y}}\) and \({\rm{z = 36 - 3}}{{\rm{x}}^{\rm{2}}}{\rm{ - 3}}{{\rm{y}}^{\rm{2}}}\) .

(b) Find the centroid of \({\rm{E}}\) (the center of mass in the case where the density is constant).

Short answer

1. Volume of the region \({\rm{E = 162\pi }}\)
2. The centroid of \({\rm{E = (0,0,15)}}\)

Step-by-step solution

To find the volume of region.

1. First, determine the point where the given paraboloids intersect: \({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ = 36 - 3}}{{\rm{x}}^{\rm{2}}}{\rm{ - 3y}}\)

To find the intersection by sorting the equation above:

\({x^2} + {y^2} = 9\)

This is a circle in the \(xy\) plane with two paraboloids as \({\rm{z}}\) coordinates. To integrate the following region in cylindrical coordinate: \({\rm{R = }}\left\{ {{{\rm{r}}^{\rm{2}}} \le {\rm{z}} \le {\rm{36 - 3}}{{\rm{r}}^{\rm{2}}}{\rm{,r}} \in {\rm{(0,3],\theta }} \in {\rm{(0,2\pi ]}}} \right\}\)

Finally, the volume includes

\(\int_{\rm{0}}^{\rm{3}} {\int_{\rm{0}}^{{\rm{2\pi }}} {\int_{{{\rm{r}}^{\rm{2}}}}^{{\rm{36 - 3}}{{\rm{r}}^{\rm{2}}}} {{\rm{rdzd\theta dr}}\int_{\rm{0}}^{\rm{3}} {\int_{\rm{0}}^{{\rm{2\pi }}} {{\rm{36r - 4}}{{\rm{r}}^{\rm{3}}}{\rm{d\theta dr = 162\pi }}} } } } } \)

Therefore, the volume of region is \({\rm{E = 162\pi }}\)

To find the mass coordinates centre.

1. If the density is constant, the total mass is calculated by multiplying the volume of portion a) by constant \({\rm{k}}\)

The total mass is transformed into \({\rm{m = 162k\pi }}\)

Then, take advantage of the fact that the region is symmetric in the plane \({\rm{x y}}\) surrounding the

circle

Because the density is constant, so deduce that

\({{\rm{M}}_{{\rm{yz}}}}{\rm{ = }}{{\rm{M}}_{{\rm{xz}}}}{\rm{0}}\)

Now figure out what occurs with the coordinate \({\rm{z}}\)

\(\begin{aligned}{{\rm{M}}_{{\rm{xy}}}}\rm &= \int_{\rm{0}}^{\rm{3}} {\int_{\rm{0}}^{{\rm{2\pi }}} {\int_{{{\rm{r}}^{\rm{2}}}}^{{\rm{36 - 3}}{{\rm{r}}^{\rm{2}}}} {{\rm{krdrd\theta dr}}} } } \\\rm &= \int_{\rm{0}}^{\rm{3}} {\int_{\rm{0}}^{{\rm{2\pi }}} {\frac{{{\rm{8kr - 216k}}{{\rm{r}}^{\rm{3}}}{\rm{ + 1295kr}}}}{{\rm{2}}}} } {\rm{d\theta dr}}\\\rm &= 2430\pi k\end{aligned}\)

Finally, the mass coordinates' centre is:

\({\rm{(0,0,15)}}\)

Therefore, the centroid of \({\rm{E = (0,0,15)}}\)