Question

While converting a triple integral from Cartesian to cylindrical coordinate system, how do I find the boundaries without drawing the graph?

Answer 1

Okay well, it would help to have an example.
But it takes a lot of algebraic manipulation.

Answer 2

You'll need a good sense of imagination and a good understanding of functions; at least.

Answer 3

You are basically manipulating inequalites

Answer 4

\[\int\limits_{-1}^{1}\int\limits_{0}^{\sqrt{1-y ^{2}}}\int\limits_{0}^{x}(x^2+y^2)dzdxdy\]

Answer 5

Okay so this is\[\begin{split}
0&<z&<x\\
0&<x&<\sqrt{1-y^2}\\
-1&<y&<1
\end{split}
\]

Answer 6

Use your transformations on these inequalities and try to solve for \(r\) and \(\theta\)

Answer 7

\[
\begin{split}
0&<z&<r\cos \theta \\
0&<r\cos\theta &<\sqrt{1-r^2\sin^\theta }\\
-1&<r\sin \theta &<1
\end{split}
\]

Answer 8

The \(z\) bounds are already done.

Answer 9

The second inequality can be squared \[
0<r^2\cos^2 \theta <1-r^2\sin^2\theta \implies 0<r^2<1 \implies 0<r<1
\]

Answer 10

Ah I see, great thanks, but what is wrong if I then transfer the inequalities to \[-1 \le y \le 1 , 0 \le x \le \sqrt{2} , 0 \le z \le \sqrt{2}\]

Answer 11

You going to go from integrating over a nice rectangular cylinder to integrating over a screwed up region in cylindrical coords.

Answer 12

The point is to make it easier, not harder.

Answer 13

Is the approach entirely wrong?