Double Integration and its Applications

**algebrapro18** · July 16th 2008, 04:14 PM

1. $\iint_{R}e^{x+y}~dA$ where R: |x| + |y| <= 2

2. A lamina occupies the part of the disk x^2 + y^2 <= 1 in the first quadrant. Find its center of mas if the density at any given point is proportional to the square of its distance from the origin.

**mr fantastic** · July 16th 2008, 04:38 PM

Originally Posted by algebrapro18

1. $\iint_{R}e^{x+y}~dA$ where R: |x| + |y| <= 2

[snip]

The region of integration in the xy-plane is bounded by the following four lines:

x + y = 2 when x > 0 and y > 0, that is, the line y = -x + 2 in the first quadrant.

-x + y = 2 when x < 0 and y > 0, that is, the line y = x + 2 in the second quadrant.

-x - y = 2 when x < 0 and y < 0, that is, the line y = -x - 2 in the third quadrant.

x - y = 2 when x > 0 and y < 0, that is, the line y = x - 2 in the fourth quadrant.

The region is therefore a diamond centered on the origin.

Life is much easier if you now make the following coordinate transformation (suggested by this region):

$u = x + y$ .... (1)

$v = x - y$ .... (2)

It follows from (1) and (2) that $x = \frac{u+v}{2}$ and $y = \frac{u-v}{2}$ . Therefore the Jacobian of the transformation is ........

The region of integration in the uv-plane is a square. The integrations are trivial.

**mr fantastic** · July 16th 2008, 04:39 PM

Originally Posted by algebrapro18

[snip]
2. A lamina occupies the part of the disk x^2 + y^2 <= 1 in the first quadrant. Find its center of mas if the density at any given point is proportional to the square of its distance from the origin.

Where are you stuck here? Show the working you've done.

**algebrapro18** · July 16th 2008, 05:19 PM

Originally Posted by mr fantastic

Where are you stuck here? Show the working you've done.

I haven't done anything I don't know how to start.

I know the center of mass:

x = $\iint_{R}xp(x,y)~dA$ / mass of the Lamina
y = $\iint_{R}yp(x,y)~dA$ / mass of the Lamina

but I don't know how to find P(x,y) and the mass of the Lamina.

**mr fantastic** · July 16th 2008, 05:32 PM

Originally Posted by algebrapro18

I haven't done anything I don't know how to start.

I know the center of mass:

x = $\iint_{R}xp(x,y)~dA$ / mass of the Lamina
y = $\iint_{R}yp(x,y)~dA$ / mass of the Lamina

but I don't know how to find P(x,y) and the mass of the Lamina.

You're given $\rho (x,y)$ !

$\rho (x,y) = k(x^2 + y^2)$ where k is the constant of proportionality.

And you should know that

$M = \int \int_{R_{xy}} \rho (x,y) \, dx \, dy$

where $R_{xy}$ is the region in the xy-plane the lamina is spread over ......

And $R_{xy}$ is also given to you. It's the part of the disk $x^2 + y^2 \leq 1$ in the first quadrant.

**algebrapro18** · July 16th 2008, 05:49 PM

I really don't feel like tying out all my work but can you check my answers?

M = (pi*k)/8
Center of mass = 8/(5*pi)

**mr fantastic** · July 16th 2008, 06:15 PM

Originally Posted by algebrapro18

I really don't feel like tying out all my work but can you check my answers?

M = (pi*k)/8
Center of mass = 8/(5*pi)

If you mean that x-coordinate = y-coordinate = 8/(5*pi) then your calculations are correct.

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