Solving to find the limits of integration for \(\phi\) when integrating with spherical coordinates

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burt

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I was working on spherical coordinates - this link goes to my last question. Now though, I'm working on this question:
1588713904161.png
I drew a picture:
1588713919897.png
And this is my work so far:

1588713976545.png

(Thank you @Dr.Peterson for helping me learn how to figure this out :))

Here is my question: I can see what \(\phi\) is from the picture - but, is there a way to numerically solve for \(\phi\)? If yes, how?
 
I have a second part of this question - my final answer was negative. Is that allowed?
 
No, a negative volume doesn't make sense here. You'll have to show your work on the integral if you can't see the error.

For [MATH]\phi[/MATH], you might consider looking just at the xz plane. What is the intersection of [MATH]z = \sqrt{x^2 + y^2}[/MATH] with that plane? What angle do the resulting lines make with the z-axis?
 
Dr.Peterson said:
No, a negative volume doesn't make sense here. You'll have to show your work on the integral if you can't see the error.

For [MATH]\phi[/MATH], you might consider looking just at the xz plane. What is the intersection of [MATH]z = \sqrt{x^2 + y^2}[/MATH] with that plane? What angle do the resulting lines make with the z-axis?
Click to expand...
1588732321140.png
This is my work. The reason why I got a negative was because of the negative cosine involved.
 
Sorry - I wanted to add this on to the last post but I couldn't edit.
The first time I went through the problem, my result was positive. But, plugging it into the calculator to check it gave me the negative result. I then went back over my work to find my error.
 
burt said:
Sorry - I wanted to add this on to the last post but I couldn't edit.
The first time I went through the problem, my result was positive. But, plugging it into the calculator to check it gave me the negative result. I then went back over my work to find my error.
Click to expand...
Did you discover that your answer is really positive? Observe that [MATH]\sqrt{2} - 2[/MATH] is negative ...

(And the cosine is not negative.)
burt said:
How can I figure out the angle from finding the point of intersection?
Click to expand...
Not from the point, but from the line.

The line I referred to is z = x. Did you get that?

The slope of that is 1. That is the tangent of the angle with the x-axis; so that angle is [MATH]\pi/2[/MATH]. And therefore so is the angle with the z axis.
 
@Dr.Peterson
Thank you!!
I think I really understand it now - the angle is just basic trig stuff. Thank you for really helping me be clear about all this! You've been incredibly helpful!
 
There's an important distinction between "a negative number" and "the negative of a number". It's good to have learned it.
 
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