Find volume between two surfaces

Question

I need to find the volume between $$x=z^2+2y^2, \space 2z+4y+x=1,$$ but I can't imagine the body I have. I tried to draw this surfaces, but it seems to me that I need something else.

Answer 1

When you are asked to find the volume between two surfaces say $z=f(x,y)$ and $z=g(x,y)$. What you do is find $\displaystyle\iint_{S}\left(f(x,y)-g(x,y)\right)\,dxdy$ where $S$ is the region enclosed by the curve $f(x,y)=g(x,y)$.

In this case you do the same but in place of $z$ you have $x$.

$$\iint_{S}(1-2z-4y-z^{2}-2y^{2})\,dydz$$

$S\equiv z^{2}+2y^{2}+2z+4y=1$

So $$(z^{2}+2z+1)+2y^{2}+4y+2=1+2+1=4$$

So $$(z+1)^{2}+2(y+1)^{2}=4$$

This is the ellipse in the $z,y-$plane

$$\frac{(z+1)^{2}}{4}+\frac{(y+1)^{2}}{2}=1$$

So you have

$$\iint_{S}(4-(z+1)^{2}-2(y+1)^{2})\,dydz=\iint_{S}4(1-\frac{(z+1)^{2}}{4}-\frac{(y+1)^{2}}{2})\,dydz$$.

Now put $u=\frac{1}{2}(z+1)$ and $v=\frac{1}{\sqrt{2}}(y+1)$.

Then $|\frac{\partial (u,v)}{\partial (z,y)}|=\begin{vmatrix}\frac{1}{2}&0\\ 0&\frac{1}{\sqrt{2}}\end{vmatrix}=\frac{1}{2\sqrt{2}}$.

So $$|\frac{\partial (z,y)}{\partial (u,v)}|=2\sqrt{2}$$

Now you are just looking at the region $u^{2}+v^{2}=1$ in the $uv-$plane which is just the circle. So by the Jacobian Transformation we have:-

$$\iint_{S'}4\cdot 2\sqrt{2}(1-u^{2}-v^{2})\,dudv $$.

Now shift to polar coordinates

$$\int_{0}^{2\pi}\int_{0}^{1}8\sqrt{2}(1-r^{2})r\,drd\theta=4\pi\sqrt{2}$$