# Double integral polar coordinates trouble

• Andrew123
In summary: Yes, that's correct. So we have the limits for z. What are the limits for r and theta?for r it would be from 0 to 3 as it's the radius of the cone and theta would be 0 to 2pi as it goes all around the coneRight! So now we have our limits:z: 0 ---> \sqrt{18-r^2}r: 0 ---> 3\theta: 0 ---> 2\piGreat! Now, since we are dealing with cylindrical
Andrew123

## Homework Statement

Consider the volume of a solid bounded by the cone: z = sqrt(x^2 + y^2) and the top half of the sphere x^2 + y^2 + z^2 = 18 that is for z >= 0

Using cylindrical coordinates, express the volume as a double integral.

## Homework Equations

easy to sketch.. we can pretty easily figure r = 3 and theta = 2 pi

## The Attempt at a Solution

I got integral of sqrt(18 - r^2) r.dr.d(theta) over r = { (rcos(theta), rsin(theta) | 0 <= r <= 3, 0 <= theta <= 2pi

but answer says... [sqrt(18-r^2) - r ] r.dr.d(theta) where have i gone wrong

thanks in advance for the help!

Remember that we are dealing with cylindrical coordinates here, *not* spherical coordinates. So r2 = x2 + y2 and the volume element is r.dr.d$\theta$ dz.

So let's try again. Can you express the equation of the cone in terms of *cylindrical* coordinates?

im so lost :( where did the other r come from?

also r.dr.d(theta).dz is triple integral not double?

Andrew123 said:
im so lost :( where did the other r come from?
Which r? The expression, $r dr d\theta dz$ is simply the volume element in cylindrical coordinates. In Cartesian coordinates, the equivalent expression would be $dx dy dz$.
Andrew123 said:
also r.dr.d(theta).dz is triple integral not double?
Indeed it is. Volume integrals *must* be triple integrals.
Andrew123 said:
Consider the volume of a solid bounded by the cone: z = sqrt(x^2 + y^2) and the top half of the sphere x^2 + y^2 + z^2 = 18 that is for z >= 0

Using cylindrical coordinates, express the volume as a double integral.
However, triple integrals may be expressed as a double integral by evaluating one of the integrals, in this case the integral with respect to z.

[sqrt(18-r^2) - r ] i mean the r bold and underlined.

So i went for z limits o to sqrt(18)

with R = {(x,y,z) in polar form | 0 <= z <= sqrt(18), 0 <=r <= 3, 0 <= theta <= 2pi}

for what integral? this is so confusing.. its just a Q for a practise exam is there anyway you could work it through abit? I find i understand much better if i can look at it to see how it works.. thankyou

Andrew123 said:
[sqrt(18-r^2) - r ] i mean the r bold and underlined.
We'll get to that eventually.
Andrew123 said:
its just a Q for a practise exam is there anyway you could work it through abit? I find i understand much better if i can look at it to see how it works.. thankyou
That's not the way it works here. I'll guide you through the question, which will help you understand it better, but I won't provide a complete or partial solution.

Can you try answering my question:
Hootenanny said:
So let's try again. Can you express the equation of the cone in terms of *cylindrical* coordinates?
In other words instead of writing

$$z = f\left(x,y\right)$$

Write it in the form

$$z = g\left(r,\theta\right)$$

so z=r for the cone in cylindrical form. now we are minusing the area from under the cone from the area under the half sphere right? so that's where we get our sqrt(18-r^2) - r from? because the z value ranges from z= sqrt(18-r^2) to z=r

Andrew123 said:
so z=r for the cone in cylindrical form.
Correct.
Andrew123 said:
now we are minusing the area from under the cone from the area under the half sphere right? so that's where we get our sqrt(18-r^2) - r from? because the z value ranges from z= sqrt(18-r^2) to z=r
Let's take it one step at a time. So, as you correct write we also have the equation of the upper half sphere:

$$z = \sqrt{18-r^2}$$

and for the cone:

$$z = r$$

Now, we need to figure out the limits. Note that the limits will be different for the sphere and cone. What are the limits for the sphere?

Sorry had to sleep.. limits for sphere would be sqrt(18-r^2) and 0 and for the cone would be sqrt(18) and 0?

Andrew123 said:
Sorry had to sleep.. limits for sphere would be sqrt(18-r^2) and 0 and for the cone would be sqrt(18) and 0?
You need to be careful here, either the upper or lower limit in each case should be a function of the radius. Your limits for there sphere are correct: the smallest value z can take is zero (at the origin) and the largest value z can take is on the surface of the sphere (i.e. z = (18-r2)1/2.

Now can you apply the same logic to the cone? What is the smallest value that z can take inside the cone. What is the largest value?

## 1. What is a double integral in polar coordinates?

A double integral in polar coordinates is a mathematical concept used to find the volume under a 3-dimensional surface in polar coordinates. It involves integrating a function over a region in the polar plane.

## 2. How do you convert a double integral in Cartesian coordinates to polar coordinates?

To convert a double integral from Cartesian coordinates to polar coordinates, you can use the following formula:
∫∫f(x,y)dxdy = ∫∫f(rcosθ, rsinθ)rdrdθ
This involves replacing x with rcosθ and y with rsinθ, and also adding an extra r term in the integral.

## 3. What is the purpose of using polar coordinates in a double integral?

Polar coordinates are often used in double integrals when the region of integration is circular or has circular symmetry. It can also make the integral easier to evaluate, as it eliminates the need to use trigonometric functions in the limits of integration.

## 4. What are common mistakes made when solving double integrals in polar coordinates?

Some common mistakes when solving double integrals in polar coordinates include forgetting to add the extra r term, using the wrong limits of integration, and not converting the function correctly from Cartesian to polar coordinates.

## 5. How can I check if my answer for a double integral in polar coordinates is correct?

You can check if your answer for a double integral in polar coordinates is correct by converting the integral back to Cartesian coordinates and evaluating it. If the value is the same as the original integral, then your answer is correct.

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