Area and Volume

Calculus II Area and Volume FRQ 1. (2011 AB3) Let R be the region in the first quadrant enclosed by the graphs of f(x) = 8x ^ 3 and g(x) = sin(pi*x) as shown in the graph. <= in(T_{1}) (1/2, 1) R y = 8y ^ 3 y = 8x ^ 3 yox? root(1/(8y), 3) = x y = sin(pi*x) arcsin(y) = pi * r' * x a. Write an equation for the line tangent to the graph of f^ * atx = 1/2 (よい) y - 1 = 6(x - 1/2) f' * (x) = 8x ^ 2 f(1/a) = 8 * (1/a) ^ 3 = 8 * 1/8 = 1 f' * (x) = 24x ^ 2 m = 6 f' * (1/2) = 24 * (1/2) ^ 2 = 274(1/4) = 6 b. int 0 ^ 1/2 ((s ln( pi < ))-8x^ 3 )dx expression in terms of x. - 1/pi * cos(pi*x) -2x^ 4 | 0 ^ n 2 (- 1/pi * cos(pi - 1/2) - Delta * (1/Delta) ^ 4) -(- 1 10 cos( 1 10 ) - 1/4 ^ 0 - 1/8 + 1/11 (1/pi - 1/5) = 193 F()-Fio) c. Find the area of R using an integral expression in terms of y. integrate (1/a * y ^ (ny) - 1/pi * s * m ^ - 1 * (y)) dy from 0 to 1 193 d. Find the volume of the solid generated by revolving R about the x - axis. J_{1} ^ (1/2) * a*pi((sin 7x) ^ 2 - (8x ^ 1) ^ 2) * dx k*pi * integrate (sin^2 (pi*x) - 64x ^ 6) dx from 0 to 1/2 5/28 * 17 = [[5], [(10pi)/28]] 561 e. Find the volume of the solid generated by revolving R about the y-axis. integrate 2pi*x(sin pi*x - 8x ^ 3) dx from 0 to U_{2} (215322 or int 0 ^ 1 pi( (1/2 * y ^ 1 * 3) ^ 2 - (1/pi plus/minus m ^ - 1 * (y)) * dxy 322 f. Set up, but do not evaluate, an integral expression to find the volume of the solid generated when R is rotated around the horizontal line y = 1 integrate a*pi(1 - y)(1/a * y ^ (4y) - 1/pi * sigma*i * n ^ - 1 * (y)) dy from 0 to 1 g. Set up, but do not evaluate, an integral expression to find the volume of the solid generated when R is rotated around the vertical line x = 1 integrate 2pi(1 - x)(sin(pi*x) - 8x ^ 3) dx from 0 to 16 f^ prime approx( (1 - sin(1x)) ^ 2 - (1 - 8 * x') * dx h. Region R is the base of solid. Cross sections of the solid axis. Find the volume of the solid. S_{n} * prod n = 1 to n ((- 1/4 * arcsin(y)) ^ 2 - (1 - 1/2 * y ^ 4) ^ 2) dy Sbh integrate (sin(pi*x) - 8x ^ 9) ^ 2 dx from 0 to 10 Calculus II Area and Volume FRQ 2. (2010 AB4) Let R be the region in the first quadrant bounded by the graph of y = 2sqrt(x) the horizontal line y = 6 and the y-axis, as shown in the graph. (9.6) y = 2sqrt(x) R 2sqrt(x) 1/2 * y = sqrt(x) (1/2 * y) ^ 2 = x 1/4 * y ^ 2 = x 0 a. Find the area of R using an integral expression in terms of x. $(6-2√x)dx = f (6-2x)dx 6x-x=(6/19)-3(9))-(0) 54-36 b. Find the area of R using an integral expression in terms of y. integrate (1/4 * y ^ 2 + 0) dy from 0 to ∞ = integrate 1/4 * y ^ 2 dy from 0 to 6 = 1 12 y^ 3 | o ^ 6 = 1 12 (6)^ 3 -0= 6*6-6 12 =(15 c. Find the volume of the solid generated by revolving R about the x - axis. integrate pi' * ((varphi) ^ 2 - (a * x ^ 14) ^ 2) dvarphi from 0 to 9 = pi * integrate (x - 4x) dx from 0 to 14 (26x-2x²) 10 pi[(36 * 9 - 2 * (9) ^ 6) - 10) ]=(162 pi d. Find the volume of the solid generated by revolving R about integrate Pi' * ((1/4 * y ^ 2) ^ a - (0) ^ 2) dy from 0 to 6 integrate 2pi(6 - 2sqrt(x)) dx from 0 to 9 = 1 16 pi* 1 5 y^ 5 | 8 ^ 6 = 1/50 * pi' * (6) ^ 5 =( 98i 0 5 pi^ prime =305.363 e. integrate y ^ x dy from a to bx Set not evaluate, an integral expression to find the volume of the solid generated when R is rotated around the horizontal line y = 6 int 0 ^ 9 mathcal pi ^ prime ( (6 - 2sqrt(x)) ^ 2 0 -(6-6)^ 2 )d varphi \mathcal{pi} * integrate (6 - 2sqrt(x)) ^ 2 dy from 0 to 4 integrate 2pi(\mathbb{E})(1/4 * y ^ 2 + 0) dy from 0 to 6 f. Set up, but do not evaluate, an integral expression to find the volume of the solid generated when R is rotated around the vertical line x = - 1 integrate 2pi(x + 1)(b - 2sqrt(x)) dx from a to a int 6 ^ 6 pi^ prime (( i 6 y^ 2 -^ prime )^ 2 - (o + 1) ^ 2 )dy g. Region R is the base of solid. Cross sections of the solid are semicircles perpendicular to the x-axis. Find the volume of the solid. integrate 1/2 * pi * r ^ 2 dr integrate 1/2 * pi * ((6 - 2sqrt(x))/2) ^ 2 dvarphi from 0 to ∞ 1/8 * pi * integrate (6 - 2x ^ 112) ^ 2 dy from 0 to 9 1/r * pi int 0 ^ 4 (36-24y^ prime prime 2 +4y)dy 1/8 * pi(36y - 16x ^ (3/2) + 2x ^ 2) | a ^ 9 1/8 * pi[(36(9) - 16 * (9) ^ (4/2) + 2 * (9) ^ 2) + 0] 21.206 1/8 * pi' * (324 - 432 + 162) = 1/8 * pi' * 0.54 * (0.27pi)/4 Calculus II Area and Volume FRQ 3. (2010B AB1) In the figure below, R is the shaded region in the first quadrant bounded by the graph of y = 4 * ln(3 - x) the horizontal line y = 6 and the vertical line x = 2 (0,6) y = 6 R 41 x = 2 y = 4 * ln(3 - x) 1/4 * y = ln(3 - x) 이 (2.0) e ^ (1/(4y)) = 3cx e ^ (1/(4y)) - 3 = - x - e ^ (1/4 * y) + 3 = x a. Find the area of R. integrate (y - 4in(z - x)) dvarphi from 0 to 2 6.817 b. Find the volume of the solid generated when R is revolved about the x - axis. integrate Pi((b) ^ 2 - (4 * ln(3 - x)) ^ 2) dchi from 0 to 2 174.463 c. Find the volume of the solid generated when R is revolved about the horizontal line y = 8 -integrate pi((8 - 4 * ln(3 - x)) ^ 2 - (8 - 6) ^ 2) dpsi from 0 to 2 168.180 d. Find the volume of the solid generated when R is revolved about the horizontal line y = - 2 integrate pi((6 + 2) ^ 2 - (48 + 4 * ln(3 - psi) + 2) ^ 2) dmu from 0 to 2 260.123 j = - 2 e. The region R is the base of a solid. For this solid, each cross section perpendicular to the x -axis is a square. Find the volume of the solid. Sbh int 0 ^ 2 ( ((6 - 4 * ln(3 - varphi)) ^ 2) dx 26.267