Cubic area of a semi circle I am not correct what am I doing wrong?

Question

Antarctica is roughly semicircular, with a radius of 2000 km (Fig. 1-5). The average thickness of its ice cover is 3000 m. How many cubic centimeters of ice does Antarctica contain? (Ignore the curvature of Earth.)

http://www.flickr.com/photos/25945858@N07/2785748122/

look at the semi circle in the photo

This is what I did

((1/2)pi r^2)*depth

so ((1/2)*pi*2000^2)*3

I come up with 18849555.9

Answer 1

(1) The area of the sem-circle is (1/2)*pi*r^2 = (1/2)*pi*2000^2 km^2

(2) The depth/thickness of 3000 m, expressed in km, is 3 km, so the

volume of ice is (1/2)*pi*4000000*3 km^3 = (1/2)*pi*1.2*10^7 km^3

(3) Now, 1 km = 1000 m = 100000 cm, i.e., 1 km = 10^5 cm, thus

(1 km)^3 = (10^15) cm^3, so . . .

(4) in cm^3, ice volume is (1/2)*pi*1.2*10^22 cm^3 = about

1.884955592*10^22 cm^3

In other words, your number is ok, but the units are not cm^3. For future problems of this type, it is a good idea - as above - to keep careful track of the units as you go.

Answer 2

Area * Depth = Volume

Area = 1/2*pi*r^2

Area 2000000*pi km^2

Thickness = 3km

Volume = 6000000*pi km^3 = 6*pi*10^6 km^3

1 km = 10^3 m = 10^5cm

1km^3 = (10^5)^3 =  10^15 cm^3

6*pi*10^6 km^3 = 6*pi*10^6 * 10^15 cm^3 = 6*pi*10^21 cm^3 18.84955592 * 10^21 cm^3

or

1.884955592 * 10^22 cm^3

or

18,849,555,921,538,800,000,000 cm^3

What you came up with was the volume in km^3.  The question asked for cm^3

Answer 3

looks like a units problem to me.  You have cubic kilometers there.