Find density of disk

[jwpaine]

A lamina occupies the part of the disk x^2 + y^2 <= 9 in the first quadrant. Density at each point given by function P(x.y) = (x^2 + y^2). Find total mass

I said:

m = $\displaystyle \int \int_{D} (x^{2} + y^{2}) dA = \int_{y=0}^{3} \int_{x=0}^{\sqrt{9-y^{2}}} (x^{2} + y^{2})dxdy$

which gives me:

m = \(\displaystyle \frac{1}{3} \int_{y=0}^{3} (9-y^{2})^{\frac{3}{2}}} dy\)

Which is not giving the right answer.

Any help would be greatly appreciated.

Thanks,
John

 

[Deleted member 4993]

A lamina occupies the part of the disk x^2 + y^2 <= 9 in the first quadrant. Density at each point given by function P(x.y) = (x^2 + y^2). Find total mass

I said:

m = $\displaystyle \int \int_{D} (x^{2} + y^{2}) dA = \int_{y=0}^{3} \int_{x=0}^{\sqrt{9-y^{2}}} (x^{2} + y^{2})dxdy$

which gives me:

m = \(\displaystyle \frac{1}{3} \int_{y=0}^{3} (9-y^{2})^{\frac{3}{2}}} dy\)
Which is not giving the right answer.

Any help would be greatly appreciated.

Thanks,
John

$\displaystyle m \, = \, \int \int \rho (x) \cdot dA$

I would do this problem in polar co-ordinate like Glen showed below..... [edit]

 

[BigGlenntheHeavy]

I am assuming that we are dealing with a planar lamina where we consider density as mass per unit surface area, not mass per unit volume.

Rectangular Coordinates: $\displaystyle m = \int_{0}^{3} \int_{0}^{\sqrt{9-x^{2}}}(x^{2}+y^{2})dydx = 81\pi/8$

Polar Coordinates: $\displaystyle m = \int_{0}^{\pi/2} \int_{0}^{3}r^{3}drd\theta = 81\pi/8$

Polar Coordinates are easier.