Posted by Kim on .
Use differentials to find the approximate change in the surface area when the radius of the sphere is increased from 10cm to 10.2cm.
I know A=4pir^2 and A' is 8pir but I'm not sure how to set it up from here. Can someone please show me all the steps involved? (This isn't HW, just a textbook question that I can't figure out, thanks!)

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Kim,
Ha! I figured it out.
dA=8pir dr
r=10 , dr=10.2
dA=8(10)(.2)pi
dA=16pi (approx)
check:
A(10)=4(10)^2pi
A(10)=400pi
A(10.2)=4(10.2)^2pi
A(10.2)=416.16
A(10.2)A(10)=16.16pi (exact)