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I have the triple integral $\iiint x dV$ where $E$ is the solid bounded by surfaces $y=x^2, x=y^2, z+x^2+y^2 = 2$, and the $xy$-plane.

What will be the cylindrical bounds for this integral? I'm pretty sure this is broken down to $\iiint(\text{from}\ z=0 \ \text{to}\ z=-x^2-y^2+2) x dzdA$, but I'm having trouble figuring out what the graph of the solid is and thus what the bounds will be.

If someone could edit my question to have the $x^2, y^2$ be the fancy type notation instead, I'd love to learn how to do this for future questions. Thanks!

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  • $\begingroup$ if you enclose x^2 + y^2 in dollar signs it will appear in what you call the 'fancy notation'. For more complicated expressions you will have to learn LaTeX. $\endgroup$ – user20266 Dec 14 '11 at 16:33
  • $\begingroup$ For latex, here's a WYSIWYG editor: codecogs.com/latex/eqneditor.php make sure you enclose the output in dollar signs. $\endgroup$ – The Chaz 2.0 Dec 14 '11 at 16:33
  • $\begingroup$ Good to know, thanks! $\endgroup$ – StudentProgramee Dec 14 '11 at 16:45
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You just have to intersect the curves $y = x^2$ and $x = y^2$ in the $xy$-plane. This will give you the limits for $x$:

$$ x = y^2 = x^4 \quad \Longleftrightarrow \quad 0 = x - x^4 = x(1-x^3) \quad \Longleftrightarrow \quad x =0,1 $$

So, the bounds for your integral are:

$$ \iiint_E xdV= \int_0^1\int_{x^2}^{\sqrt{x}}\int_0^{2-x^2-y^2}xdzdydx \ . $$

Hint: draw both curves in the $xy$-plane to understand the limits for $y$.

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youre integrating over the solid with $0\leq z\leq 2-x^2-y^2$ ie between the $xy$-plane and a downward facing paraboloid. in the $xy$-plane, you are integrating over the region enclosed between the two parabolas $y=x^2, x=y^2$, ie between $y=x^2$ and $y=\sqrt{x}$ for $0\leq x\leq 1$ (or between $x=y^2, x=\sqrt{y}$ for $0\leq y\leq 1$). always draw a picture!!! so something like $$ \int_0^1\int_{x^2}^{\sqrt{x}}\int_0^{2-x^2-y^2}x\ dzdydx\ \ {\rm or }\ \ \int_0^1\int_{y^2}^{\sqrt{y}}\int_0^{2-x^2-y^2}x\ dzdxdy $$

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A good first step is to convert the $z$ bound to cylindrical coordinates.

As you mentioned, you can pull the $x^2+y^2$ to the other side of the upper-bound equation, giving you $z = 2 - (x^2+y^2)$. This can be converted to $z = 2 - r^2$, as $x^2+y^2 = r^2$ in cylindrical coordinates.

Therefore, you know that the $z$-bounds will be $\int_{0}^{2-r^2}$.

Next, you want to draw the shadow of the graph in the $xy$ plane. By tracing $r$-rays over the shadow, you can see that the $r$ limits will be from $0$ to the point where $x^2 = \sqrt{x}$, which is $1$. Now, trace $\theta$-rays to find the minimum and maximum of $\theta$. You can see that $\theta$ goes from $0$ to $\pi/2$

Finally, you want to convert $dV$ to cylindrical coordinates. When converting from rectangular to cylindrical, $dV$ becomes $rdzdrd\theta$.

What we're left with is:

$$\int_{0}^{\pi/2}\int_{0}^{1}\int_{0}^{2-r^2}r^2cos(\theta)dzdrd\theta$$

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