Question

Find the area of the region between curves

1.   

2.   

1. Find Find the area of the region between curves by rotating about x-axis the region in the x,y- plane bounded below and above, respectively, by the curves: a. y = 2x2, y = 4x + 16 b. x = -y2 + 10, x = (y – 2) I

Answer 1

a) 2 y - 4+1 Solution the errea between Curves is the area between a curve f(x) and a Curve ja) interval [a, b] A-s'agca) din Here, f(x) and 962) - 4% +16 To find intersection point [a, b] f(x) g(x) 2x² a 42 +16 9.2 4x -16 - 0 2X-8 = (-4) (x+2) = 240 x 4 2 2. = -2,6-4 Now Area, 1) 4 3. -16% 4x 2 5x] 2 1122 12x-1x +16)1 day 3-2 (4) ² - 16x [22 1945-1604). Les 2.st [128 -32-64)-( 4 – 8 +32) (032-94) - (14 +24) - 128 +1 5) 128 +16 - 94-24 → Area is alway rive The area between the curves y y = = 4% +16 is 72 sq. ->/128 120 - 72 144 3 Hence, and

- y² +10 and xo (y-2) Solution't b) x=- the area between curves in area between a curve f(x) and g(x) on an interval [a,b 16 A. - Jolyon-goa) / dz xk X -S G X: +10 -x tlo y² y . y = $w-x + 10 r-x+10 and Iax +2 N-ztlo or youw x=(y-2)2 y-z = idx ago yi fwling intersection paintin 2 + 4x or 2-rx N-xflo = -1-x+10 FX +10 + -x+10= 2x tlo - (x = 10 N-x+10 = x + 2 1-xflo de +2 No solution for XER retio dx + 2 No solution for XER N- xt10 = x = 9 dx + 2 = -x + 2 x = 0 de +2 Area = 15412 - 6-*a * 2)fali Sis*10-474 + 2)da 1.89-07 5x+10 -1- xt1o|dx

lisa-2-68 ) Asia 233 pelada Ex]. 4 3 56 and 9/0 16 si -N-x tlo 99 9 F2710-(-5x + 2)/dx xlo dx - s/dx + 2) dx $2 - (*) - 5 -v-7410 lx 220 W-x+10 Toldx = 28 (w-xt to Ide let, -x+10=4 Sdu du 21-41. skudu)) 326-44) [{u}]. § +5 Hence, The area between the curves ye to and xa(y-2) is 64 sq. unit 9 - 2 2 t-s-rudu) = 2 22x2 it helfs 3 4 3 Hole Please rate Now, 4 64 3 3