# Tagged Questions

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### Set up triple integral for volume (cylindrical coordinates)

I am given the following question Let $D$ be the region in $\mathbb{R}^3$ that lies within $x^2 + y^2 =4$, underneath the surface $z= 4- x^2 - y^2$ and above the surface $z=- \sqrt{9-x^2 - y^2}$ ...
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### Finding the Limits of the Triple Integral (Spherical Coordinates)

Let $D$ be the region in $\mathbb{R}^3$ below $z=-\sqrt{x^2 + y^2}$ and above $z=-\sqrt{4-x^2 -y^2}$. Rewrite \begin{align*}\iiint \limits_D z^2 dV\end{align*} using Spherical Coordinates. I rewrote ...
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### Marginal density function question

The question and answer is shown but I don't fully understand the answer for part a. Could someone please explain to me why the integral setup for the marginal density function of y1 is from y1 to 1, ...
### Find the centroid of the boomarang shaped region for the parabolas $y^2=-4(x-1)$ and $y^2=-2(x-2)$
I know the formulas, I only need assistance setting up the initial integral. So my order of integration must be $\mathrm{d}x$ $\mathrm{d}y$. Then if we solve the parabola for $x$ the new integral we ...
### Use the transformation $x=u^2-v^2$, $y=2uv$ to evaluate the integral
$$\int_0^1 \int_0^{2\sqrt{1-x}} \! \sqrt{x^2+y^2} \, \mathrm{d}y\,\mathrm{d}x$$ Here's where I'm at: $J(x,y)=4u^2+4v^2$ Substituting $x$ and $y$ into the integral: \$\sqrt{(u^2-v^2)^2+4u^2v^2} ...