Thread: Finding mass of solid region

1. Finding mass of solid region

Can anyone help. I think I set the integral up properly but I'm not sure I'm taking the actual integral correctly because I keep getting really complicated numbers.

Find the mass of the solid region bounded by the parabolic surfaces z=16-2x^2-2y^2 and z=2x^2+2y^2 if the density of the solid is sqrt(x^2+y^2)

I set it up as follows;

integral from x=-2 to 2 , integral from y= -sqrt(4-x^2) to sqrt(4-x^2) , integral from z= 2x^2+2y^2 to z=16-2x^2-2y^2 of (sqrt(x^2+y^2)) dzdydx

(sorry I don't know the symbols so that looks awful)

2. I assume you are also bounded by the cylinder $\displaystyle x^2+y^2=4$?

(To see how you type an equation, click on one of mine).

Did you try integrating in cylindrical coordinates? Seems like it would be pretty easy then.

$\displaystyle r^2=x^2+y^2$

3. Ok so we have a region bounded by:

$\displaystyle z = 16-2 { x}^{ 2} -2 { y}^{ 2}$
$\displaystyle z = 2 { x}^{ 2} +2 {y}^{ 2}$

At what z do they cross? You can show that they cross at z = 8 (try x = 2, y = 0). Okay so now we can do the upper and lower parts separately. We have:

Mass = $\displaystyle 2 \int_{ - 2 }^{ 2 } \int_{- \sqrt[ ]{4- {x }^{ 2} } }^{ \sqrt[ ]{4- {x }^{ 2} } } \int_{ 8}^{ 16- 2{ x}^{ 2}-2 { y}^{ 2} } \sqrt[ ]{ { x}^{2 }+ { y}^{2 } } dzdydx$

This is horrible, lets do cylindrical coordinates instead as suggested.

$\displaystyle \int_{0 }^{2 } \int_{0 }^{2 \pi } \int_{ 2 {r }^{ 2} }^{ 16-2 { r}^{ 2} } { r}^{ 2} dzd \theta dr$

Much cleaner! Notice in the r^2 one of the r's comes from the square root of r^2, and the other from the jacobian (the r that you need for cylindrical coordinates).

I got $\displaystyle \frac{512 \pi }{ 15}$ units for a final answer.

4. I don't understand how to transform into cylindrical coordinates...???

5. I assume that you have done that section? What don't you understand?

The basic premise is that you can represent the coordinates of a point in different ways. You can think of it as (x,y,z), or you can represent the point in cylindrical coordinates (r,$\displaystyle \theta$,z) (polar coordinates with z), or in spherical $\displaystyle (r,\theta,\phi)$

In your example, the object is easily represented as a bunch of thin cylinders (all with axis along the z-axis), i.e. there is a nice rotational symmetry about the z-axis, and so cylindrical coordinates is probably a good way to go. If the object had rotational symmetry around the origin, you might pick spherical.

How can we tell this object has nice symmetry about the z-axis? Well for starters, notice all the instances of $\displaystyle x^2+y^2$. Circle right?

So first let's describe the volume of integration using cylindrical coordinates.

r:0->2
$\displaystyle \theta$:0->2$\displaystyle \pi$
So far we just described the equivalent of x:-2->2, y:$\displaystyle -\sqrt{4-x^2}$->$\displaystyle \sqrt{4-x^2}$

Now z:$\displaystyle 2x^2+2y^2$->$\displaystyle 16-2x^2-2y^2$
But we need to write this in terms of r and $\displaystyle \theta$.
But, as in polar coordinates, $\displaystyle r^2=x^2+y^2$ so:
$\displaystyle 2x^2+2y^2=2(x^2+y^2)=2r^2$
$\displaystyle 16-2x^2-2y^2=16-2(x^2+y^2)=16-2r^2$

z:$\displaystyle 2r^2$->$\displaystyle 16-2r^2$

Now do you see where the limits of integration came from in Mentia's post?

2 final pieces. First we have to convert the integrand into (r,$\displaystyle \theta$,z).

Easy. $\displaystyle \sqrt{x^2+y^2}=\sqrt{r^2}=r$ (technically, we took the positive sqrt since densities should be non-negative.

Finally, if you just integrated with $\displaystyle dz\,d\theta\,dr$ you would not get the right answer. The problem boils down to the fact that the volume $\displaystyle \Delta z\,\Delta\theta\,\Delta r$ does not equal $\displaystyle \Delta x\,\Delta y\,\Delta z$. Or you can think of it as you are doing an integration by substitution and so you have to do the change on the differential part of the integral too (if you set u = x^2 to do some integral, then $\displaystyle du=2x\,dx$. So the formula you just have to know that the $\displaystyle dx\,dy\,dz"="r\,dr\,d\theta\,dz$, or in our case, because of the way we set up the integral, $\displaystyle r\,dz\,d\theta\,dr$

6. Integral Questions

Can anyone help.

integral from 4 to 1
x=( x^2-x+1)/(sqrt(x))

integral from pie to pie/4
x=secxtanx

integral from 4 to x
u=(2+(sqrt(u))^8

integral
x=sqrt(1+x^2x^5)

integral
x=tan^6xsec^4x

7. Originally Posted by colaname
Can anyone help.

integral from 4 to 1
x=( x^2-x+1)/(sqrt(x))

integral from pie to pie/4
x=secxtanx

integral from 4 to x
u=(2+(sqrt(u))^8

integral
x=sqrt(1+x^2x^5)

integral
x=tan^6xsec^4x
colaname, this post does not belong in this thread. Only one problem (even if multipart) should be in a thread. Please go back to the forum and post your questions there. Thanks.