Greetings,(adsbygoogle = window.adsbygoogle || []).push({});

I somewhat see the explanation but I need a conversation to clear the fog.

Part of a larger problem is this integral:

Find the volume of a cone [tex] r = 3sin(theta) [/tex] inside the sphere [tex] x^2 +y^2 + z^2 = 9 [/tex]

[tex]

\int \int \int r \,dz dr dtheta

[/tex]

My limits:

z goes from 0 to [tex] \sqrt{9-r^2}[/tex] with the integral multiplied by “2” to handle the +/- side of the square-root

r goes from 0 to 3 sin(theta)

and initially I had theta going from 0 to 2pi.

First integration:

[tex]

\int \int r\sqrt{9-r^2}\, dr dtheta

[/tex]

r from 0 to 3sin(theta)

theta from 0 to 2pi – again initially – I reasoned that this is not correct

Second integration:

[tex]

-\frac{2}{3} \int (9-r^2)^{2/3}\, dtheta

[/tex]

r from 0 to sin(theta)

[tex]

-\frac{2}{3} \int (9-9sin(theta)^2)^{2/3} - -9^{3/2}\, dtheta

[/tex]

With my limits from 0 to 2pi – wrong answer – I said “Ah the area under a sinusoidal – so 2 * the same integral from 0 to pi”

Wrong again

I then plot the polar plot of r = 3 sin(theta) and make a table and see that at pi/2 I get the first half of the “circle” polar plot.

So I said, “Ah 4* the integral from 0 to pi/2 to get all the way back to 2pi” – see I had 2pi on the brain.

Again – wrong answer, so I see that the plot completes the “circle” polar plot when the angle gets to pi.

So when I multiply the integral by 2 and go from 0 to pi/2 I get it.

Can you help clear the fog?

When I see an integral of a sinusoid, my first instinct was to multiply the integral by 2 and then integrate from 0 to pi.

I assume this is “Cartesian-thinking?”

Next, I want to say when working with polar coordinates, that the angle should “sweep” all the way around from 0 to 2pi.

I know the answer is associated with the plot of “r” – should one always generate a plot to see how “r” goes or is there a more straightforward way to set up the integral?

Thanks

Sparky

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# Setting up polar-integrals

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