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Well, the book tells me to dray the set:

$$x^2+y^2\le a^2, y^2+z^2\le a^2, (a> 0)$$

So I interpreted this as the volume of something under the region $x^2+y^2\le a^2$. This something should be the value of $z$, which I did: $z = \sqrt{a^2-y^2}$

I tried to integrate then like this:

$$\int_{-a}^a \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\sqrt{a^2-y^2}dy \ dx $$

so I would be integrating $z = \sqrt{a^2-y^2}$ from the bottom of the circumference of radius $a$, to its top, depending on the chosen $y$. Then I would integrate this for all $y$, therefore, from the 'beggining' of the circumference to its 'end', that is, from $-a$ to $a$. This should calculate the volume desired. Is it wrong? Because Wolfram Alpha doesn't even try to calculate it.

Poperton
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  • Wait, is the question to draw the set, or to compute it's volume? Or it's surface area, because volume would be a triple integral. Anyway, if it's just the first, note that there are $3$ variables, so we must be in at least $3D$, and also note that $x^2 + y^2 \le a^2$ is a solid cylinder, as is the second region. The intersection of these is then not hard to visualise. – stochasticboy321 Oct 24 '15 at 00:44
  • I should calculate the volume of the set generated by these constraints – Poperton Oct 24 '15 at 00:47
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    http://math.stackexchange.com/questions/494249/find-the-volume-common-to-two-circular-cylinders-each-with-radius-r-if-the-axe should answer your question. Also, this in case you want more exposition on it. – stochasticboy321 Oct 24 '15 at 01:04

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