Let E be the region bounded between the graphs of y=5-x^2, y=0, and x=2. Let S b
ID: 3284025 • Letter: L
Question
Let E be the region bounded between the graphs of y=5-x^2, y=0, and x=2. Let S be the solid generated by rotating the region E about the x-axis. a) Sketch the region E showing a representative vertical rectangle and the disk(washer with no hole)traced out by this rectangle. Just give a rough sketch of the three dimensional disk. b)Express symbolically the volume V of the solid S as a limit of a sum. Assume n washers of equal width are used that the radii are determined by the right endpoints xi of the subintervals. Final answer should not contain the letter x or any other letter but n or i. 1) What is the width deltax of each washer? 2) What is the volume of deltaVi of the ith washer? Do not simpify. 3) What is the symbolic expression for the volume of the solid S expressed as the limit of a sum? Use part 2 results but do not simplify. 4) Find exact volume of the solid S using integration and the Fundamental theorem of Calculus. Answer should be a fraction multiplied by pi... I am trying but cant find any examples to help me through so I have no clue what to do or start...Explanation / Answer
Answer:- i hope it would help you 1) Try washers: each washer will have an outer radius R = secx and an inner radius r = 0, so V = ?[a,b] p(R² - r²) dx = p?[0,p/3] sec²x dx = p tanx |[0,p/3] = p(tan(p/3) - tan(0)) V = p(v3 - 0) = pv3 ? choice 4 2) or "about the line x = -3". Let's try shells. Each shell will have radius R = 3 + x and height h = y2 - y1 = x - x² and thickness dx, so V = 2p?[0,1] (3 + x)(x - x²) dx = 2p?[0,1] (3x - 2x² - x³) dx V = 2p((3/2)x² - (2/3)x³ - (1/4)x^4) |[0,1] = 2p(3/2 - 2/3 - 1/4) = (2p/12)(18 - 8 - 3) V = (p/6)(7) = 7p/6 ? choice 4 3) Since both curves are in terms of y, let's try washers: Each washer will have an outer radius R = 1 + x1 = 1 + vy and an inner radius r = 1 + x2 = 1 + y² and thickness dy, so V = ?[a,b] p(R² - r²) dy = p?[0,1] ((1 + 2vy + y) - (1 + 2y² + y^4)) dy V = p?[0,1] (2y^(1/2) + y - 2y² - y^4) dy = p((4/3)y^(3/2) + ½y² - (2/3)y^3 - (1/5)y^5) |[0,1] V = p(4/3 + ½ - 2/3 - 1/5) = (p/30)(40 + 15 - 20 - 6) = 29p/30 4) We have to use washers here because the curve crosses the y-axis at y = 1. Then y = e^(x/2). Each washer will have an outer radius R = y = e^(x/2) and an inner radius r = 0 and thickness dx, so V = ? p(R² - r²) dx = p?[0,6] e^x dx = pe^x |[0,6] = p(e^6 - e^0)
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