Integral Calculus.

Hi can you help me solve this with a solution and explanation. There's the example after the question for your reference. Thank you so much J

QUESTION: Is it possible to get the area between y2 = 3x and y = x3, using a. vertical strips b. horizontal strips Why? or Why not? Area Between Curves Add Slide Method 1 (Vertical Strip): f(x ) Area = [f(x) -(g(x))dx Ig (x) upper lower h= f (x ) - 9 ( x ) function function Method 2 (Horizontal Strip): Area = [f ( ) (g(y)]dy + dy right left function function Area Between Curves x interpt , x = - 4 y- intercept, y-+12 Add Slide xinteeptix- . Example: y- intercept , y = 3/ 3 Find the area of the region bounded by the equations x = 3y - 2 and x = 2y2 - 4 x= sy - 2 4 - 242 Using horizontal strip : 2 - 42 * = 24 -y Area - ( " [ fcy ) - 9 ( y ) ] dy dy 2/3 Determining y.8 42 Area - J . " [ ( By - 2 ) - ( 24 -4 ) ] dy - 1/2 yi = 2 X=X 3 y-2 = 242-4 = [ (- 2y 2 + 3 y + 2 ) dy fly ) = * = 3y- 2 242- 3y -2 =0 gly) = x = 2424 = -24 + 34 +2y - Q.F. YI= - Z 2 1 /2 Factor yz = 2 125 unit 2 24 Using vert, Strip Area = 2 A, + A ze Add Slide -7/ 2 = 2 s + 2 ) dx + [( 8 + 2 ) "= ( 3 *+ 3 ) ]dx ye 125 unit ? 24