I have a maths assignment and i can't figure out a problem and i'm hoping someone can help me with it Consider the Region R in the first quadrant enclosed by the y axis and the graphs of y= cosx and y = sinx a) Sketch the region R b) Find the volume of the solid obtained by rotating R about the y axis c) Find the volume of the solid obtained by rotating R about the line y= -1 I know im suppose to use either the disc method or the shell method but i just cant see how the question works if anyone can help Thanks in advance
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You should know how y1 = sin(x) and y2 = cos(x) looks like. The region R will be the first intersection from the 2 curves. To find the point at that curve you set the 2 equation equal to each other. sin(x) = cos(x) divide cos(x) from both sides you get sin(x) / cos(x) = 1 OR tan(x) = 1 if you're farmiliar with trigonometry, you should know right away from sin(x) = cos(x). x = (45°)(2π/360°) or π/4 if not do inverse tan(1) in radian mode. The region R is the region made from the points (0,0), (0,1) and (π/4 , sin(π/4)) Now use shell method to find the volume of the region for part b. Shell approximation is lim as n->∞ summation of 2πxif(xi)Δx as i goes from 1 to n. So you basically get integral of 2πxf(x)dx with endpoints xi to xn. Since the region R is result from 2 curves, the volume of the region will be the subtraction of the top curve and the bottom curve. In general, let f(x) = y1 = sin(x) and g(x) = y2 = cos(X). V = 2π∫xg(x)dx - 2π∫xf(x)dx = 2π∫x[g(x) - f(x)]dx from xi to xn From the intersection we know the xi = 0 and the xn = π/4. V = 2π∫x[cos(x) - sin(x)]dx Use integration by parts ∫UdV = UV - ∫VdU Let U = x, dV = [cos(x) - sin(x)]dx. Then dU = dx and V = [sin(x) + cos(x)] ∫UdV = x[sin(x) + cos(x)] - ∫[sin(x) + cos(x)]dx ∫UdV = xsin(x) + xcos(x) + cos(x) - sin(x) V = 2π[xsin(x) + xcos(x) + cos(x) - sin(x) from 0 to π/4] V = 2π[[(π/4)sin(π/4) + (π/4)cos(π/4) + cos(π/4) - sin(π/4)] - [0 + 0 + 1 + 0]] sin(π/4) = cos(π/4) Simplifies to: V = 2π[(2π/4)cos(π/4) - 1)] V = π²cos(π/4) - 2π ≈ 0.696 For part c, you need to push the 2 functions up by unit of 1 since the revolution axis is 1 below 0 and make the x-axis the revolution axis. f(x) = sin(x) is now f(x) + 1 = sin(x) + 1. g(x) = cos(x) is now g(x) + 1 = cos(x) + 1 where g(x) > f(x) for 0≤x<π/4. Now that the revolution axis is on the x-axis we can use the disk method. In general: V = ∫A(x)dx = π∫f(x)²dx where A(x) is the cross section area of the disk. V = π∫[(g(x)+1)² - (f(x) + 1)²]dx from xi to xn From the intersection we know the xi = 0 and the xn = π/4. V = π∫[(cos(x) + 1)² - (sin(x) + 1)²]dx Factoring out: V = π∫[(cos²(x) + 2cos(x) + 1) - (sin²(x) + 2sin(x) + 1)]dx V = π∫(cos²(x) + 2cos(x) - sin²(x) - 2sin(x))dx Trigonometry identity: cos²(x) - sin²(x) = cos(2x) V = π∫(cos(2x) + 2cos(x) - 2sin(x))dx V = π[sin(2x)/2 + 2sin(x) + 2cos(x) from 0 to π/4] V = π[(sin(π/2)/2 + 2sin(π/4) + 2cos(π/4)) - (0 + 0 + 2)] sin(π/4) = cos(π/4) V = π[1/2 + 4cos(π/4) - 2] ≈ 4.17 I think those are the answers
