Volume of solid $W$ delimited by $z=x^2+3y^2$ and $z=8-x^2-y^2$. Volume of solid $W$ delimited by $z=x^2+3y^2$ and $z=8-x^2-y^2$.
I need to solve this with triple integrals, however I'm having a lot of problems. Anyone knows how to solve?
 A: The hard part of such problems is to imagine the volume enclosed by the surfaces and describing the points inside the volume in a mathematical language so that you can determine the limits of integration. 
It seems that the two surfaces are some parabolic surfaces facing toward each other. You may imagine some picture in your mind like the figure below
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ 
So, first we find the intersection of the two surfaces as it bounds the variation of $x$ and $y$ coordinates of the points inside this volume. Hence, consider the following
$$\eqalign{
  & {x^2} + 3{y^2} = 8 - {x^2} - {y^2}  \cr 
  & 2{x^2} + 4{y^2} = 8  \cr 
  & {{{x^2}} \over 4} + {{{y^2}} \over 2} = 1  \cr 
  & {\left( {{x \over 2}} \right)^2} + {\left( {{y \over {\sqrt 2 }}} \right)^2} = 1 \cr} $$
and hence all points $(x,y,z)$ which lie on both surfaces satisfy
$$\matrix{
   {\left\{ \matrix{
  {\left( {{x \over 2}} \right)^2} + {\left( {{y \over {\sqrt 2 }}} \right)^2} = 1 \hfill \cr 
  z = {x^2} + 3{y^2} \hfill \cr}  \right.\,\,\,\,\,or\,\,} & {\left\{ \matrix{
  {\left( {{x \over 2}} \right)^2} + {\left( {{y \over {\sqrt 2 }}} \right)^2} = 1 \hfill \cr 
  z = 8 - {x^2} - {y^2} \hfill \cr}  \right.}  \cr 
 } $$
which are two different equations of the same intersection curve. Now, we are ready to describe the coordinates of the points inside the volume in some mathematical language
$$\Omega  = \left\{ {\left( {x,y,z} \right)| - 2 \le x \le 2, - \sqrt {2 - {1 \over 2}{x^2}}  \le y \le \sqrt {2 - {1 \over 2}{x^2}} ,{x^2} + 3{y^2} \le z \le 8 - {x^2} - {y^2}} \right\}$$
and hence your triple integral will be
$$\eqalign{
  & \mathop{\int\!\!\!\int\!\!\!\int}\limits_{\kern-5.5pt \Omega } 
 {dV}  = \int\limits_{x =  - 2}^2 {dx\int\limits_{y =  - \sqrt {2 - 0.5{x^2}} }^{\sqrt {2 - 0.5{x^2}} } {dy\int\limits_{z = {x^2} + 3{y^2}}^{8 - {x^2} - {y^2}} {dz} } }  = \int\limits_{x =  - 2}^2 {\int\limits_{y =  - \sqrt {2 - 0.5{x^2}} }^{\sqrt {2 - 0.5{x^2}} } {\int\limits_{z = {x^2} + 3{y^2}}^{8 - {x^2} - {y^2}} {dzdydx} } }   \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int\limits_{x =  - 2}^2 {\int\limits_{y =  - \sqrt {2 - 0.5{x^2}} }^{\sqrt {2 - 0.5{x^2}} } {\left[ {\left( {8 - {x^2} - {y^2}} \right) - \left( {{x^2} + 3{y^2}} \right)} \right]} } dydx  \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int\limits_{x =  - 2}^2 {\int\limits_{y =  - \sqrt {2 - 0.5{x^2}} }^{\sqrt {2 - 0.5{x^2}} } {\left( {8 - 2{x^2} - 4{y^2}} \right)dydx} }   \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4\int\limits_{x =  - 2}^2 {\int\limits_{y = 0}^{\sqrt {2 - 0.5{x^2}} } {\left( {4 - {x^2} - 2{y^2}} \right)dydx} }   \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4\int\limits_{x =  - 2}^2 {\left[ {\left( {4 - {x^2}} \right)y - {2 \over 3}{y^3}} \right]} _0^{\sqrt {2 - 0.5{x^2}} }dx = 4\int\limits_{x =  - 2}^2 {\left[ {\left( {\left( {4 - {x^2}} \right) - {2 \over 3}{y^2}} \right)y} \right]} _0^{\sqrt {2 - 0.5{x^2}} }dx  \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4\int\limits_{x =  - 2}^2 {\left[ {\left( {{8 \over 3} - {2 \over 3}{x^2}} \right)\sqrt {2 - 0.5{x^2}} } \right]dx}   \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{32} \over 3}\int\limits_{x = 0}^2 {\left[ {\left( {2 - 0.5{x^2}} \right)\sqrt {2 - 0.5{x^2}} } \right]dx}   \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{32} \over 3}\left( {{{3\sqrt 2 \pi } \over 4}} \right) = 8\sqrt 2 \pi  \cr} $$
A: Let's first find the boundary of the integration region, that is, where both surfaces intersect:
$z=x^2+3y^2=8-x^2-y^2$ thus $2x^2+4y^2=8$ thus it's the ellipse $\frac{x^2}{2^2}+\frac{y^2}{\sqrt 2^2}=1$ in the $xy$ plane.
Now let $E$ be the plane region bounded by that ellipse. The volume of $W$ is then $\displaystyle\iint_E\left(8-x^2-y^2-(x^2+3y^2)\right)\,dx\,dy=\iint_E\left(8-2 x^2-4 y^2\right)\,dx\,dy$.
In order to define the integration limits one should change coordinates as follows: $x=2\,r\cos\theta$, $y=\sqrt2\,r\sin\theta$, so that the bounding ellipse is simply $r=1$, $0\leq\theta\leq2\pi$ in the new coordinates. Now let's calculate the Jacobian:
$\displaystyle J=\begin{vmatrix}\frac{\partial x}{\partial r}&\frac{\partial x}{\partial \theta}\\\frac{\partial y}{\partial r}&\frac{\partial y}{\partial \theta}\end{vmatrix}=\begin{vmatrix}2\cos\theta&-2\,r\sin\theta\\\sqrt2\sin\theta&\sqrt2\,r\cos\theta\end{vmatrix}=2\sqrt2\,r$.
Therefore, the volume is
$\displaystyle\iint_E\left(8-2 x^2-4 y^2\right)\,dx\,dy=\int_0^1\int_0^{2\pi}\left(8-2\cdot4r^2\cos^2\theta-4\cdot2r^2\sin^2\theta\right)2\sqrt2\,r\,d\theta\,dr=8\cdot2\sqrt2\int_0^1(1-r^2)r\,dr\int_0^{2\pi}d\theta=8\cdot2\sqrt2\frac14\cdot2\pi=8\sqrt2\pi$.
