Question

8. Let E be the solid in the first octant bounded by: the plane 2x + y + z = 8, the vertical cylinder y = x2, and the coordinate planes x = 0 and z = 0. For each of the three parts below you must illustrate your solution with diagrams in 2 and 3 dimensions. Marks will be given for the quality of the diagrams and how they are able to help the reader understand the way in which the solid is being cut up. Your diagrams must be hand-drawn. (a) [4 marksCalculate the volume V of E using an iterated integral of the form S s dzdydx dzdydx You must show your calculations.

Answer 1

Na 72n+1+2=8 yax2 - X Dark portion in the figure is the region E. We now find the point of intersection of an+y+z=8 and y=x2 in Zzo plome in the 1st octant. :. 20+ y 28 and y=x2 => x²+2x-8=0 => x² + ax=2x-820 =>(x+4) (x-2)=0 . x=2. 1 on it is in the first octomr). :. a) Volume v of E ssl dz dy dx. r from the diagram we see that to find the volume of the region , z vorries from 720 plane to 2n + Y + Z28 plane , then y verries from yox? to the line 2x+y +2=8 in the plane Z=0 ta varies from o to 2 . (ie the point p . 2 8-21 8-4-2x V= dzdy du no 2 8-22 42 of S**[27:-*-** ay din IL (8-y-24 ) dy du . 8-2n (8-9-2x ) dy du AL

1-24 -(-2018)(8 -za) – (8-2x)2 – (-2418) ?(? du -5. [68-22) 2-.(3-2) ? (8 - 24) 2+ + 23% - S. (1.68-243°– 82 +223 + 4x^ ] in -$ (264-x)?-8x2+2x3+4** ) dan = $." (22 –16x + 2x?-82? +2m?+ 4 x 9 ) dom - 1,(1 +2?=6x²-162732 ) du 3 - = 2 + -2.2-802% +32.2 +8 – 16-32 +64 Ve 134 3) b) vs Is dy dz de 2 8 8-3-2n I dydzda from the figure ly varies from the cylinder y=x2 to the plane 2x+y + Z28 4 then Iz vorries from a to 8 & then a from o to 2 ] . c) v IS s dy dx dz so se dy du dz 8-3-24 (on here only change the logic is also same as (b), is a verries before z.