alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

# Y iy by applying the pythagoras ip theorem with p as

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Unformatted text preview: 1 32109876543210987654321 32109876543210987654321 32109876543210987654321 32109876543210987654321 32109876543210987654321 32109876543210987654321 32109876543210987654321 32109876543210987654321 32109876543210987654321 32109876543210987654321 32109876543210987654321 32109876543210987654321 32109876543210987654321 32109876543210987654321 32109876543210987654321 32109876543210987654321 32109876543210987654321 32109876543210987654321 32109876543210987654321 32109876543210987654321 r R (0,0) The elements of area are rings similar elements in shape but varying in size. x element of area dA = 2πrdr R R area of circle A = ∫dA = ∫ 2πrdr = [ πr 2 ] = πR2 0 0 y R } d arc = Rdθ 10987654321 10987654321 10987654321 10987654321 10987654321 10987654321 10987654321 10987654321 10987654321 10987654321 10987654321 dθ (0,0) The elements of area a re triangular e lements wedges similar in shape and equal in size. θ / / element of area dA = 1 2(Rdθ).R = 1 2R2dθ area of ciricle A = ∫dA = 2π ∫ 1 2R2dθ / 0 190 2 2π = [ R θ ] = πR2 20 R x volume of a cylinder volume Example Ex ample 9: f(x) (0 , 0) 6543 654321 654321 543212109876543210987654321098765654321 4321 543212109876543210987654321098765654321 43212109876543210987654321098765432121 4321 543212109876543210987654321098765654321 5 4321 radius = r rdθ f(x) = r dx height = h x The “area under the cur ve” f(x) = r over the interval [0, h] is : “area curv “ar h h h ∫ 0 f (x) . d x = ∫ 0 r d x = [ r x ]0 = r . h This is the area of a cross-section. If we rotate this rectangular cross-section area r . h thru an infinitesimal angle dθ around the x-axis, we will get a wedge shaped block of volume dV. volume of wedge dV = 1/2 (cross-section area . r dθ) = 1/2 r2 . h . dθ We may now rotate this wedge dV from θ = 0 to θ = 2π radians to get the solid cylinder of height h and radius r. volume of cylinder V = ∫ 2π dV = ∫ 1 2r2 h. d θ / 0 191 2π 2 = [ r h θ ] = π r2 h 20 volume of a cone volume Example 10: Ex ample 10: rx = /h f(x) dx 54321 54321 54321 54321 54321 54321 54321 54321 54321 54321 (0 , 0) dx heig...
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