How would I solve question 1?

The region R in the first quadrant is bounded by the y-axis, y = 6, and the graph of y = x2 + 2. Which integral would find the volume of the solid generated when R is revolved around the line y = 6. ON A T ( x 2 + 2 ) 2 dx B ) | 7: 1 6 - ( x 2 + 2 ) 1 2 dx 6 / T ( x 2 + 2 ) 2 doc o D) It [6 - (x2+ 2) 12 dx EX / T: [ 6 - ( x 2 + 2 ) 1 2 dx 1 11.180 2. Which of the following gives the area of the region bounded by f (x) = =x3 + =x2 - 6x + 2, and g(x) = ex - 7 from x = 0 to x = 2? 1.12 A ) | ( 5 x 3 + 7 x 2 - 6x + 2 ) - (ex - 7) | dx+ [ [ ( ex - 7) - (5x3 + 2x2 - 6x+ 2) dx o 1.12 703 B ) / ( 5 x 3 + 7 8 2 - 62 + 2 ) - (ex - 7) | dx + [ [ ( ex - 7 ) - ( 52 3 + 7 2 2 - 6* + 2 do . 703 .703 8 / [ ( ex - 7) - (3 x 3 + 4 x 2 - 6* + 2) | + [ [( 2x3+782-6*+2) -(ex-7)| dx .703 D ) J [ ( x 3 + 2 7 2 - 6x + 2 ) - (ex - 1 / dx