Triple Integral: Volume Between z+x^2+y^2=4 and x^2+y^2+z^2=6

[Kuma]

Homework Statement

find the volume between

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

Homework Equations

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?

 

[ehild]

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?

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

 

[vela]

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

 

[Kuma]

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

 

[vela]

The cap is in between the two surfaces.

 

[queenstudy]

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

but what if i want to make z variable doesn't 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 ??

 

[vela]

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.

 

[queenstudy]

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