The region bounded by the curve y = x² + 1 and the line y-x+3 is revolved about the x-axis to generate a solid. Find the volume of this solid. Solution: x² + 1 = x + 3 x²+x-2=0 (x+2)(x-1) = 0 x=-2, x = 1 Outer radius: R(x) = x+3 Inner radius: r(x) = x² + 1 = L" = ([R(x)]² - [r{(x)]³) dx V = (-2,5) R(x)--x+3 y--x+3 (1,2) (-x+3)² - (x² + 1)²) dx = 」7(=x = 7(8 6x - x²-x4) dx Interval of integration (a) i need clear ans and solve very very fast in 20 min and thank you | DYBALA ROX) ra (1, 2)) 8x 3x²------² - 117 = Washer cross section Outer radius: R(x)--x+3 Inner radius: r(x) = x²+1 (b) H.W. 1: Solve the previous example by rotating the region about the line y = -1 ?

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