Polar coordinate bounds for double integral over hemisphere and disk

[Mdhiggenz]

Homework Statement

Step 1) I put the following into polar coordinates
√(16-x²-y²)=√16-r²

Where r≤4

step 2 I solved for y in the original problem which is in the link

y≤√(4-x²)

step 3. I graphed the above function

step 4. I put the above function in polar coordinates getting

r≤2

So I already had an idea that the graph would be from

0≤r≤2 dr However my d(theta) was incorrect I got that it would be 0≤theta≤∏ as my interpretation of the graph shows. The answer however shows 0≤theta≤2∏

Which does't make much sense to me. Where did I got wrong?

Homework Equations

The Attempt at a Solution

 

[tiny-tim]

what is the actual question?

 

[Mdhiggenz]

Upper left hand corner of the graph, the problem statement says Use a multiple integral and a convenient coordinate system to find the volume of the solid. I forgot to type it up I apologize.

 

[tiny-tim]

i still don't get it

x²+ y² ≤ 4 isn't a circle, it's a solid cylinder

anyway, show us how you found the line of interection​

 

[Mdhiggenz]

Honestly I don't know I solved as if it was a circle.