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A: I am attaching image so that you understand each and every step.
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Find the area of the surface obtained by rotating the curve about the x-axis
plz answer 5,10
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- 1- The given curve is rotated about the y-axis. Find the area of the resulting surface. y = 1/3x3/2, 0 ≤ x ≤ 12 2- The given curve is rotated about the y-axis. Find the area of the resulting surface. x = (a2 -y2 )1/2 , 0 ≤ y ≤ a/9 3- If the infinite curve y = e−7x, x ≥ 0, is rotated about the x-axis, find the area of the resulting surface. 4- Use Simpson's Rule with n = 10 to approximate the area of the surface obtained by rotating the curve about the x-axis. Compare your answer with the value of the integral produced by a calculator. (Round your answers to six decimal places.) y = 5xex, 0 ≤ x ≤ 18.2) 13) Find the exact area of the surface obtained by rotating the curve about the xaxis.9.3.16. Compute the surface area of the surface obtained by revolving the given curve about the indicated axis. (a) about the x-axis (b) about x = 4 please answer both a and b 1 to t to 2 x=4t, y=sqrt(t^2) but if you can only do one please do b
- 1) Find the area of the surface generated by rotating the function y=x^3 about the x-axis over 0≤x≤3 2) Find the area of the surface generated by rotating the function g(y) = (9-y^2) ^1/2 about the y-axis over 0≤y≤2.Find the area of the surface generated when the curve y=1/4(e2x + e−2x)is rotated about the x-axis from x = -2 to x = 2.5. Find the area bounded by the parabolas y2=4x and y2+12x=36 6. Find the area bounded by the curve y=4x-x2and the lines x=-2 and y=4 7. Find the area bounded by the functions y=3x-x3and line y=2 8. Find the area in the first quadrant bounded by the curve x2y=a3and the lines x=2a, y=4a, and the axes.
- Find the area enclosed by the line y = x − 3 and the parabola y2 = 2x + 18. SOLUTION By solving the two equations we find that the points of intersection are , −4 and , 6 . We solve the equation of the parabola for x and notice from the figure that the left and right boundary curves are xL = 1 2 y2 − 9 and xR = y + 3. We must integrate between the appropriate y-values, y = −4 and y = 6. Thus A = 6 (xR − xL) dy −4 = 6 (y + 3) − dy −4 = 6 dy −4 = 6 −4 = + 18 + 72 − + 8 − 48 = .1. Find the equation of the tangent plane to the surface z= x2sin(xyπ/2) +3y at (-1,1,2). 2. Find the equation of the tangent plane to the surface z=ln(1+xy) at the point (1,2ln3). While I believe that I have found the answers to these problems I want to doule check my solutions: the answer to 1. is: 2xsin((πxy)/2)+((π2ycos((πxy)/2))/2)*(x+1)+((πx3cos((πxy)/2))/(2)+3)*(y-1)-(z-2) the answer to 2. is: ((2ln3)/1+2ln3)*(x-1)+(1/1+2ln3)*(y-2ln3)+ln(1+2ln3)You are given a cylinder S in R3 with equation x = z2 + 2. Let C be the curve in the xz-plane whose equation is the same as that of S. Find the equation of the surface of revolution generated by revolving C about the x-axis.
- Consider the surface x^4+ 3xz + z^2+ cos( πxy ) = -2 and the point P0 ( -1, 1, 2) on that surface. Find an equation of (a) the tangent plane at P0 (b) the normal line to the surface at P0Consider a curve represented by x2 + y2 = 4x.(a) Find the polar equation of f.(b) Set up the integral to find the area inside the curve f and outside r = 2.use Taylor’s formula for ƒ(x, y) at the origin to find quadratic and cubic approximations of ƒ near the origin.1. ƒ(x, y) = xe^y 2. ƒ(x, y) = e^x cos y3. ƒ(x, y) = y sin x 4. ƒ(x, y) = sin x cos y5. ƒ(x, y) = e^x ln (1 + y) 6. ƒ(x, y) = ln (2x + y + 1)7. ƒ(x, y) = sin (x2 + y2) 8. ƒ(x, y) = cos(x2 + y2