How do you find the area under the arch of Sin^3(x)Cos^3(x)?

Dr Tran

New member
I've been stressing over this issue and I'm rather stuck. I'll need work shown...but my answer seems very off. The limits of integration are [0,(pi/2)]
pi*r^2 is the area formula so wouldn't it end up as (Sin^6(x)Cos^6(x))?
I'm feeling stupid because I've been doing this like a Volume problem rather than area. Thank you for your help!
 
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