Question

Calculus III. HELP PLEASE

Answer 1

Find the volume in the first octant bounded by x^2+y^2=a^2 and z=x+y? Show all of your work

Answer 2

Great! What work will you be showing?

Can we use Polar Coordinates? That might make parts of it a little easier?

Really, though, since we get to use the 1st Octant, making EVERYTHING positive, I'm tempted to solve the first equation for 'y'.

Answer 3

I don't know how to do it with polar coordinates yet. And the only thing I can think for is setting up the integral like this:
\[\int\limits_{0}^{\infty}\int\limits_{0}^{\infty} f(x,y) dx dy \]
I don't know what to put in for \[f(x,y)\] because I have two equations....

Answer 4

I would use cylindrical coordinates\[x=r\cos\theta\]\[y=r\sin\theta\]\[z=z\]

Answer 5

\[\int\limits_{0}^{\pi/2}\int\limits_{0}^{\infty} \] right? How do I write what I'm actually integrating though?

Answer 6

Have you considered "x+y" or 'z', depending on how you set it up...

Also, [0,Unbounded]? I don't think so. You've a circle. Try [0,a]

Answer 7

so \[\int\limits_{0}^{\pi/2} \int\limits_{0}^{a}\]?
and what i would integrate is
\[(r \cos(\theta)+ r \sin(\theta))*r dr d(\theta)\]

Answer 8

Yes, but it would have been more instructive to set up your integral as\[V=\iiint dV=\int_0^{\pi/2}\int_0^a\int_0^{r\cos\theta+r\sin\theta}rdzdrd\theta\]that makes it clear you ar finding a volume. after the first evaluation you will get what you have of course

Answer 9

Right. Unless you have been COMMANDED to use a double integral, please use the triple integral as this ENTIRELY avoids the coufusion concerning the argument inside the integral expression.

Answer 10

I have been commanded to use a double integral.
So the final integral would be...
\[\int\limits_{0}^{\pi/2}\int\limits_{0}^{a} (r \cos(\theta)+ r \sin(\theta))*r dr d(\theta)\]
right?

Answer 11

Go with confidence.