Uplift Education



466725-212725Applications Of The Definite IntegralApplications Of The Definite IntegralThe area between two curvesThe Volume of the Solid of revolution (by slicing)1. AREA BETWEEN the CURVES dA=outer?function-inner?function dx2476501231900032385002095500A=abdA=aby1x-y2(x)dx A=cddA=cdx1y-x2(y)dyEX: Determine the area of the region bounded by y = 2x2 +10 and y = 4x +16 between x = – 2 and x = 5 350520018478500 A=abdA=abouterfunction-innerfunctiondx-21907514605100EX: Determine the area of the region enclosed by y = sin x and y = cos x and the y -axis for 0≤x≤π2 319087580645000444500EX:? Determine the area of the enclosed area by x=12y2-3 and y=x-1 32480252095600Intersection: (-1,-2) and (5,4).95250742950033299403937000? THE SAME: Determine the area of the enclosed area by x=12y2-3 and y=x-1 3457575660390029527519240500So, in this last example we’ve seen a case where we could use either method to find the area.? However, the second was definitely easier.4324350160020002209800-14478000EX: Find area Intersection points are: y = - 1y = 3 A=-13-y2+10-y-22dy=-13-2y2+4y+6dy=-23y3+2y2+6y3-1 A=643Volume of REVOLUTION? Find the Volume of revolution using the disk method? Find the volume of revolution using the washer method? Find the volume of revolution using the shell method? Find the volume of a solid with known cross sectionsArea is only one of the applications of integration. We can add up representative volumes in the same way we add up representative rectangles. When we are measuring volumes of revolution, we can slice representative disks or washers. DISK METHOD V=abdV dV=πr2dx 011049000dV=πfx2dxdV=πfx+k2dxdV=πfx-k2dxdV=πk-fx2dx1-19040076200027622500V=cddVWASHER METHODA solid obtained by revolving a region around a line.-123825205105000-190500857256350000Volumes by Cylindrical Shells0122555000-4445000-1905000-38100018097520193000The Volume for Solids with Known Cross Sections377190012192000Procedure: volume by slicing sketch the solid and a typical cross section find a formula for the area, A(x), of the cross section find limits of integration integrate A(x) to get volumeFind the volume of a solid whose base is the circle x2 + y2 = 4 and where cross sections perpendicular to the x-axis are a) squares 22098009525 x2 + y2 = 4 y=4-x2 length of a side is : 24-x2 dV?=?A?dx??????????A?=a2?V?=?4-224-x2dx =128300 x2 + y2 = 4 y=4-x2 length of a side is : 24-x2 dV?=?A?dx??????????A?=a2?V?=?4-224-x2dx =1283-95251206500b) Equilateral triangles 220980010160 x2 + y2 = 4 y=4-x2 A=?12a?a2-a22=34?a2=34-?x2 V=-2234-?x2dx=323≈18.475020000 x2 + y2 = 4 y=4-x2 A=?12a?a2-a22=34?a2=34-?x2 V=-2234-?x2dx=323≈18.4757620015684500235267585725 x2 + y2 = 4 y=4-x2 A=?12?π?a22=18π?a2=?π?4-x22 dV=A?dx???? V=-22π?4-x22dx=16π3≈16.75500 x2 + y2 = 4 y=4-x2 A=?12?π?a22=18π?a2=?π?4-x22 dV=A?dx???? V=-22π?4-x22dx=16π3≈16.755 c) semicircles d) Isosceles right triangles254317547625x2 + y2 = 4 y=4-x2 A=?12a?a2tanπ4=?a24=4-?x2 dV=A?dx???? V=-224-?x2?dx=323≈10.66700x2 + y2 = 4 y=4-x2 A=?12a?a2tanπ4=?a24=4-?x2 dV=A?dx???? V=-224-?x2?dx=323≈10.667PRACTICE: 1. Find the volume of the solid generated by revolving about the x-axis the region bounded by the graph of fx=x-1 the x-axis, and the line x = 5. Draw a sketch. 1. ANS: 8π 2. Find the volume of the solid generated by revolving about the x -axis the region bounded by the graph of y=cosx where 0≤x≤ π2 the x-axis, and the y-axis. Draw a sketch. 2. ANS: π3. Find the volume of the solid generated by revolving about the y-axis the region in the first quadrant bounded by the graph of y = x2, the y-axis, and the line y = 6. Draw a sketch. 3. ANS: 18 π 4. Using a calculator, find the volume of the solid generated by revolving about the line y = 8 the region bounded by the graph of y = x 2 + 4, the line y = 8. Draw a sketch. ANS: 512/15 π 5. Using a calculator, find the volume of the solid generated by revolving about the line y = –3 the region bounded by the graph of y = ex, the y-axis, the lines x = ln 2 and y = – 3. Sketch. 5. ANS: 13.7383 π 6. Using the Washer, find the volume of the solid generated by revolving the region bounded by y = x 3 and y = x in the first quadrant about the x-axis. Draw a sketch. Method (just a fancy name – use sketch and common sense!!! instead of given boundaries, you have to find it as intersection of two curves and then use sketch to subtract one volume from another ) 6. ANS: 4π/217. Using the Washer Method and a calculator, find the volume of the solid generated by revolving the region bounded by y = x 3 and y = x about the line y = 2. Draw a sketch. 7. ANS: 17π/21 8. Using the Washer Method and a calculator, find the volume of the solid generated by revolving the region bounded by y = x2 and x = y2 about the y-axis. Draw a sketch. 8. ANS: 3π/1041814758191500AGAIN PRACTICE:1. The base of a solid is the region enclosed by the ellipse x24+ y225 =1 . The cross sections are perpendicular to the x-axis and are isosceles right triangles whose hypotenuses are on the ellipse. Find the volume of the solid. 1. ANS: V = 200/3 2. The base of a solid is the region enclosed by a triangle whose vertices are (0, 0), (4, 0) and (0, 2). The cross sections are semicircles perpendicular to the x-axis. Using a calculator, find the volume of the solid.2724150301625002. ANS: V = 2.094 3. Find the volume of the solid whose base is the region bounded by the lines x + 4 y = 4, x = 0, and y = 0, if the cross sections taken perpendicular to the x-axis are semicircles. 3. ANS: V = π/6 4. The base of a solid is the region in the first quadrant bounded by the y-axis, the graph of y = arctanx, the horizontal line y = 3, and the vertical line x = 1. For this solid, each cross section perpendicular to the x-axis is a square. What is the volume of the solid? 4. ANS: V = ∫?? (3 - arctan(x))? dx = 6.61233 370522574295005. A solid has its base is the region bounded by the lines x + 2y = 6, x = 0 and y = 0 and the cross sections taken perpendicular to x-axis are circles. Find the volume the solid.5. ANS: 9/2 π 6. A solid has its base is the region bounded by the lines x + y = 4, x = 0 and y = 0 and the cross section is perpendicular to the x-axis are equilateral triangles. Find its volume. 6. ANS: V = 163/3 ................
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