# Let f(x,y)=1-x^2/4-y^2 and Omega={(x,y) in mathbb{R}^2:f(x,y) geq 0}. Compute the volume of the set A={(x,y,z) in mathbb{R}^3:(x,y) in Omega, 0 leq z leq f(x,y)}

Evaluating the Volume of a Cupola-Shaped Set by Integration
Let $f\left(x,y\right)=1-\frac{{x}^{2}}{4}-{y}^{2}$ and $\mathrm{\Omega }=\left\{\left(x,y\right)\in {\mathbb{R}}^{2}:f\left(x,y\right)\ge 0\right\}.$
Compute the volume of the set $A=\left\{\left(x,y,z\right)\in {\mathbb{R}}^{3}:\left(x,y\right)\in \mathrm{\Omega },0\le z\le f\left(x,y\right)\right\}.$
My idea is to slice the set along the z-axis, obtaining a set ${E}_{z}$ - in fact, an ellipse - and computing the volume as ${\int }_{0}^{1}{\int }_{{E}_{z}}dxdydz$.
However, I am stuck finding a way to describe ${E}_{z}$. What is the best strategy to do that?
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ahem37
Step 1
Use this parameterization for the whole 3 dimensional space
$\begin{array}{rl}x& =2u\mathrm{cos}v\\ y& =u\mathrm{sin}v\\ z& =w\end{array}\phantom{\rule{2em}{0ex}}0\le u<\mathrm{\infty },\phantom{\rule{1em}{0ex}}0\le v<2\pi ,\phantom{\rule{1em}{0ex}}-\mathrm{\infty }
then the integral for volume will be
$V={\int }_{v=0}^{2\pi }{\int }_{u=0}^{1}{\int }_{w=0}^{1-{u}^{2}}\frac{\mathrm{\partial }\left(x,y,z\right)}{\mathrm{\partial }\left(u,v,w\right)}dw\phantom{\rule{thinmathspace}{0ex}}du\phantom{\rule{thinmathspace}{0ex}}dv$
Step 2
Where $\frac{\mathrm{\partial }\left(x,y,z\right)}{\mathrm{\partial }\left(u,v,w\right)}=|\begin{array}{ccc}\frac{\mathrm{\partial }x}{\mathrm{\partial }u}& \frac{\mathrm{\partial }x}{\mathrm{\partial }v}& \frac{\mathrm{\partial }x}{\mathrm{\partial }w}\\ \frac{\mathrm{\partial }y}{\mathrm{\partial }u}& \frac{\mathrm{\partial }y}{\mathrm{\partial }v}& \frac{\mathrm{\partial }y}{\mathrm{\partial }w}\\ \frac{\mathrm{\partial }z}{\mathrm{\partial }u}& \frac{\mathrm{\partial }z}{\mathrm{\partial }v}& \frac{\mathrm{\partial }z}{\mathrm{\partial }w}\end{array}|=|\begin{array}{ccc}2\mathrm{cos}v& -2u\mathrm{sin}v& 0\\ \mathrm{sin}v& u\mathrm{cos}v& 0\\ 0& 0& 1\end{array}|=2u$ is the jacobian.