** Will medal **
Calc 3
Can someone show me how you would covert the limits of integration?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- rsst123

** Will medal **
Calc 3
Can someone show me how you would covert the limits of integration?

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- rsst123

##### 1 Attachment

- ganeshie8

what do you know about the surfaces
\(z=\sqrt{x^2+y^2}\)
\(z=\sqrt{8-x^2-y^2}\)
?

- rsst123

one is a cone and the other is a sphere correct?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- ganeshie8

Yes, something like this ?
|dw:1436972486503:dw|

- rsst123

yes

- ganeshie8

good, setup the bounds for \(\theta, \phi, \rho\)

- rsst123

i understand that θ is 2pi, but i cant figure out how to get ϕ and ρ. I know p is the radius and I think ϕ is the angle between the +z axis but i dont see it

- Empty

There are multiple ways to describe the angle \(\phi\). Just imagine the earth as being a sphere of constant radius. You can go around the equator from \(0\) to \(2 \pi\) which corresponds to our \(\theta\) value like you understand but we have two very common ways to talk about the azimuthal angle \(\phi\) which make sense.
We can start \(\phi\) from the north pole and call that \(\phi = 0\) and rotate down to the equator \(\phi = \frac{\pi}{2}\) and then to the south pole \(\phi = \pi\)
The other way that's commonly done is they'll start with the equator \(\phi = 0\) and say going up to the north pole is \(\phi = \frac{\pi}{2}\) and then say going down to the south pole is \(\phi = - \frac{\pi}{2}\)
No method is better than the other, just whichever is most convenient to your problem solving. Generally mathematicians choose the first one and physicists choose the second, but it doesn't matter.

- rsst123

wow that way makes to much sense so ϕ would be 0 to pi/4 ?

- rsst123

and what about p?

- Empty

Yeah in the first coordinate system yep. =) What's your second question asking?

- rsst123

how would i determine the radius p?

- Empty

Ahhh ok sorry I didn't actually read your question I just saw that you were confused about \(\phi\) and thought I'd help with that give me a sec haha

- rsst123

ok thanks!

- Empty

You can determine \(\rho\) from the awesome picture @ganeshie8 made, give it your best guess and try to give a reason for why you think that.
If not, tell me what the integral would be if we replaced \(x^2+y^2+z^2\) in the integral with a \(1\).

- rsst123

sorry for the late response I had to take my calc final but the information you gave helped me answer the question on my final. Thanks guys!

Looking for something else?

Not the answer you are looking for? Search for more explanations.