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November 29th, 2012, 04:09 AM   #1
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Area/Volume Problem with a cup

Let R be the region above the x-axis and below y=(1/2)x+2 where 0 <= x <= 1 together with the region above y=2sqrt(x-1)and below y=(1/2)x+2 where 1 <= x<= 10. A glass cup is created by revolving R about the x-axis where x and y are measured in centimeters.
a. how much liquid can the cup hold?
b. what volume of glass is needed to create the cup? Give your answer in cm^3
c. The final step in creating the cup is to cover it in a liquid protective coating. If the coating will cover every side of the cup (top,bottom, inside, and outside), and 1 milliliter of coating can cover 10cm^2 area, how much of the liquid protective coating is needed for the cup? Give your answer in mL.
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November 29th, 2012, 04:23 AM   #2
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Re: Area/Volume Problem with a cup

This is the region R to be revolved about the x-axis to create the glass cup:

[attachment=0:ay5q2cp0]glasscup.jpg[/attachment:ay5q2cp0]

a) Using the disk method, how would you set up an integral representing the capacity of the cup?

b) Using the disk and washer methods, how would you find the volume of glass used to make the cup?

c) What do you think is a good strategy for finding the complete surface area of the cup?
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File Type: jpg glasscup.jpg (13.1 KB, 152 views)
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November 29th, 2012, 04:31 AM   #3
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Re: Area/Volume Problem with a cup

c) I'm thinking break it into parts and add up the area. the bottom is a circle. The top is a circle - a circle. The outside is a cone with the top cut off (from 1 to 10). The inside is the difficult part.
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November 29th, 2012, 04:44 AM   #4
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Re: Area/Volume Problem with a cup

Yes, and for the inside you will need to use the method for finding a surface of rotation.
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