how to solve this question step by step **Problem 9.8** - Determine the moment of inertia of the crosshatched area about the x-axis by integration.**Description of the Diagram:**The diagram represents a crosshatched area within the first quadrant of the Cartesian plane defined by two curves: - The upper boundary is defined by the curve \( y = 9 - x^2 \).- The lower boundary is \( y = 3x - x^2 \).**Key Features:**- Both curves intersect the y-axis at \( y = 9 \) and \( y = 0 \), respectively.- The intersection of the curves occurs at \( x = 3 \), along the x-axis.- The entire crosshatched area is bounded from \( x = 0 \) to \( x = 3 \).**Axes and Dimensions:**- The vertical axis (y-axis) is labeled, with a height of 9 meters.- The horizontal axis (x-axis) is shown, extending from \( x = 0 \) to \( x = 3 \) meters.- Dashed lines indicate the dimensions of the bounded area, showing a width of 3 meters and a height reaching up to 9 meters at its peak.The purpose of the exercise is to use integration to find the moment of inertia about the x-axis for the enclosed crosshatched region.

Solution:- Given data:- To find:- The moment of inertia of x axisCalculation:- Now, IxA = IxA+B -IxB So,To find1) IxA+B y = 9 - x2x2 = 9-yx = 9-ydifferentiate wrt ydIx = dAy2 = x dy y2dIx =9-ydy y2∴ ∫09dIx = ∫09y2 9-ydytaking limits of both side and we get,IxA+B = 1166435 = 333.257Calculation:- Now, To find,2) IxB :-dIx = dAy2 = 3-2xdy y2but, y = 3x -x2∴ dy = 3-2xdx ∴ y2 = 9x2+x4-6x3dIx = 3-2x9x2+x4-6x33-2xdx∫03dIx =∫039+4x2-12x9x2+x4-6x3dxtaking limit∴ IxB = 72970 = 10.414therefore, IxA = IxA+B - IxB IxA = 333.257 -10.414 IxA = 322.842 m4Conclusion:- The moment of inertia of x axis = 322.842 m4