Lisa0gxmelicole
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Let R be the region in the first quadrant enclosed by the graph y=(square root(6x+4)), the line y=2x, and the y-axis. . . a. Find the area of R.. b.Set up but do not integrate an integral,expression in terms of a single variable for the volume of the solid generated when R is revolved about the x-axis .. c. do the same as part b, just this time the y-axis.

Answer :

Area of R is the difference between the two functions.
 Let's check for signs to find out which is the larger at x = 0: 

√(6x + 4) = 2 when x = 0 
2x = 0 when x = 0 

Thus we need to find the area of 

√(6x + 4) - 2x 

= int from 0 to 2 (√(6x + 4) - 2x)dx 


=[√(6x + 4)]^3/9 - x^2 from 0 to 2 

=[√(16)]^3 - 2^2 - {[√(4)]^3 - 0} 

= 64 - 4 - 8 

= 52 sq units 

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