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Find the volume of the solid bounded by the following: x = 1+y^2 y = x-3 Rotated

ID: 3192998 • Letter: F

Question

Find the volume of the solid bounded by the following: x = 1+y^2 y = x-3 Rotated about the y-axis

Explanation / Answer

pi * int (a to b) of (f(x))^2 dx about x-axis pi * int (c to d) of (x)^2 dy about y-axis (1) Find limits f integration 1-x^2=0 at x=+/- 1 V = pi* int(-1 to 1) of (1-x^2)^2 dx V = pi * int (-1 to 1) of 1-2x^2+x^4 dx V = pi [x - 2/3 x^3 +x^5 / 5] evaluated from -1 to 1 plug in 1 subtract plug in -1 pi[(1-2/3+1/5) - (-1 + 2/3 +1/5) ] pi [ 8/15 +8/15] = 16/15 * pi (2) solve for x= in order to plug into formula y=ln x ===> x = e^y V = pi*int{1to2} of (e^y)^2 dy V = pi *[ (1/2) e^(2y) ] eval from 1 to 2 V = pi/2 * [e^4 - e^2] (3) Find intersection by setting functions to each other 1/4 x^2 = 5 - x^2 x^2 = 20 - 4x^2 5 x^2 = 20 x^2 = 4 x = +/- 2 these are limits of integration We will do outside vlume minus inside volume V = pi * int{-2 to 2} [ (5-x^2)^2 - (1/4 x^2)^2 ] dx = pi int {-2 to 2} [25 - 10x +x^4 - 1/16 x^4 ] dx = pi [ 25x -5x^2 + 3/16 x^5 ] val from -2 to 2 = pi [(50 - 20 + 6)-(-50-20-6) = 112 pi

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