# Triple integral.

1. Dec 19, 2011

### Kuma

1. The problem statement, all variables and given/known data

find the volume between

z +x^2 + y^2 = 4 and x^2+y^2+z^2 = 6

2. Relevant equations

3. The attempt at a solution

what i did first was solve for the intersection points of z
i got 2 and -1.

then you get two equations for x^2 + y^2

x^y+y^2 = 5 and x^2+y^2=2

so then i used polars and i figure that r goes from 2 to 5, z goes from 2 to -1 and pi goes from 0 to 2pi

is that right?

2. Dec 20, 2011

### ehild

The bounds for z are correct, but the bounds of r=sqrt(x2+y2) change with z:
$$\sqrt{4-z}<r<\sqrt{6-z^2}$$

ehild

3. Dec 20, 2011

### vela

Staff Emeritus
I've attached a plot of the two surfaces. Do you also have to take into account the cap?

#### Attached Files:

• ###### plot.png
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4. Dec 20, 2011

### Kuma

No, just the portion in between. Thanks for the plot.

5. Dec 20, 2011

### vela

Staff Emeritus
The cap is in between the two surfaces.

6. Dec 23, 2011

### queenstudy

but what if i want to make z variable doesnt r become constant???
and one more thing when we you are finding the domain in cartesian is polar coordinates also considered to be in the cartesian way ??

7. Dec 23, 2011

### vela

Staff Emeritus
Yes, you can let r run between constant values and have the limits for z depend on r, if you want. Ehild's suggestion, however, will make the math simpler.

8. Dec 23, 2011

### queenstudy

i get it 3 domains for z if i look at the figure i get it thanks