Setting up a Double Integral in Polar Coordinates

[Drakkith]

Homework Statement

Consider the 'ice cream cone' bounded by
z =

14 − x2 − y2 and z =

x2 + y2
.(a) Find the equation of the intersection of the two surfaces in terms of x and y.
(b) Set up the integral in polar coordinates.

Homework Equations

The Attempt at a Solution

I got part a without any trouble. You just set each equation equal to one another and get:
$x^2+y^2=7$

I also found my limits of integration just fine: $\int_0^{2\pi} \int_0^{\sqrt{7}} rdrd\theta$

I just can't seem to set up the integrand correctly. I thought it would be: $x^2+y^2-7 = (rcos(\theta))^2+(rsin(\theta))^2 - 7 = r^2 (cos^2(\theta)+sin^2(\theta)) - 7 = r^2-7$

That would make the integral: $\int_0^{2\pi} \int_0^{\sqrt{7}} (r^3-7r) drd\theta$

However that's not correct and I can't seem to find the correct one.

 

[SammyS]

Homework Statement

Consider the 'ice cream cone' bounded by
z =

14 − x2 − y2 and z =

x2 + y2
.(a) Find the equation of the intersection of the two surfaces in terms of x and y.
(b) Set up the integral in polar coordinates.

Homework Equations

The Attempt at a Solution

I got part a without any trouble. You just set each equation equal to one another and get:
$x^2+y^2=7$

I also found my limits of integration just fine: $\int_0^{2\pi} \int_0^{\sqrt{7}} rdrd\theta$

I just can't seem to set up the integrand correctly. I thought it would be: $x^2+y^2-7 = (rcos(\theta))^2+(rsin(\theta))^2 - 7 = r^2 (cos^2(\theta)+sin^2(\theta)) - 7 = r^2-7$

That would make the integral: $\int_0^{2\pi} \int_0^{\sqrt{7}} (r^3-7r) drd\theta$

However that's not correct and I can't seem to find the correct one.

I suppose those equations should read

$z = \sqrt{ 14 − x^2 − y^2\ }\$ and $\ z = \sqrt{ x^2 + y^2\ }$ .

What is it that integrating is supposed to give you ?

 

[Drakkith]

I suppose those equations should read

$z = \sqrt{ 14 − x^2 − y^2\ }\$ and $\ z = \sqrt{ x^2 + y^2\ }$ .

Whoops, I meant to fix those before I posted...

What is it that integrating is supposed to give you ?

Some sort of ice cream cone shape. Perhaps I'm supposed to integrate using those other two equations instead of where they intersect?

 

[LCKurtz]

You need to integrate$$
\iint_R z_{upper} - z_{lower}~dA$$where $R$ is the region you have figured out, and the $z$'s are expressed in terms of $r,\theta$.

 

[SammyS]

(b) Set up the integral in polar coordinates.

My question could have been more specific.

Are you integrating to find surface area, or to find volume?

 

[Drakkith]

My question could have been more specific.

Are you integrating to find surface area, or to find volume?

I'm pretty sure it's volume. We haven't done anything with surface area yet.

You need to integrate$$
\iint_R z_{upper} - z_{lower}~dA$$where $R$ is the region you have figured out, and the $z$'s are expressed in terms of $r,\theta$.

Ah, of course. It's just like Calc 2 where you find the region between two lines, except this time it's 2 surfaces.

So that should be:
$\int_0^{2\pi}\int_0^\sqrt{7} (\sqrt{14-r^2} - r) rdrd\theta = \int_0^{2\pi}\int_0^\sqrt{7} (r\sqrt{14-r^2} - r^2) drd\theta$
Because $x^2+y^2 = r^2$, so $\sqrt{14-x^2-y^2} = \sqrt{14-(x^2+y^2)} = \sqrt{14-r^2}$
And $\sqrt{x^2+y^2} = \sqrt{r^2} = r$

Now to integrate...