Question

Gen Ed Assessment for MATH 2414 Calculus II

 

Important Submission Instructions to the Student:

1. All explanations must be typed. The supporting math work must be neat and suitable for copying or scanning. The final document should be saved and submitted as a single

.pdf file.

2.  Give the assignment the heading GEN ED ASSESSMENT: MATH 2414.

3.  Do not include your name in the assignment or in the file name.

4.  Submit your assignment to your instructor via Blackboard by the deadline established by your instructor.

 

 

Gen Ed Assessment for MATH 2414 Calculus II:

 

 

 

Let R be the region in the first quadrant bounded by the graphs of

Let S be the solid formed by revolving the region R about the x-axis.

 

Complete the items below. Include enough detail that a Calculus I student would be able to follow each step of your work. Please use correct mathematical notation and label graphs appropriately.

 

1.      Sketch the two dimensional region R. Label each boundary of the region with a corresponding equation.

2.      Set up an integral with respect to x representing the volume of the solid S. Identify which method you used (disk/washer or shell) and evaluate this integral.

3.      Set up an integral with respect to y representing the volume of the solid S.

Identify which method you used (disk/washer or shell) and evaluate this integral.

4.      Compare and discuss the two integrals, addressing each of the questions below. In your writing, imagine that your audience is your Calculus II classmates.

a.      Which method would you recommend to a fellow student? Why?

b.     What errors might a student make when using an integral in terms of x?

c.      What errors might a student make when using an integral in terms of y?

d.     Present and work an example of a problem that would be better solved using the other method. (The one that you did not recommend in part a.)

Answer 1

(3 ) Region bounded by the curves y=x+1, y = 3+1, X=0 is rotated about the y-axis To find the volume of solid generatel y =+ Intersection points of two curves. y = 3+1 = x+7 - 3--6=0 (x-3)(2+2)=0 = 3, 2=-2 (not applicable) y=(38²+1 = 10 Interrecting point (3,10) (i) Now set up the integral with respect tos. we use shell method radius of shell = x height of shell = (x+7) - (x²+1) = x+7-0²| = x-x²+6 Volume of solid - 27/(radius) height dac = 2Pacax-x+6) ds = at ſlzex2+62)dx = 2* [ + 332 = 2*( - +27] -0 = 27 ( 9-20.astar) - 21 (1575) = 31.57 Volume of roolid = 31.5t cubic units=98.96 Cubic (11) Set up integral by disk method when integral with respect to y. We will find the volume of solid by only curve y=x²+1 rotated about y-axis from y=1 to y=10. And volume of solid by only Curve y=x+7 rotated about y-axis from y=7 toy=10 Then difference of volume is the requined volume of solid.

1o cij Volume of solid rotate the curve y=x+1 about yaxis radius of dust = x area of disk = T22 = F(Y-1) Volume = 5 764 -1) dy = T(-y); = *[C100 - 10) - (+ -)] = [40-6-08] = 40.5 ñ cin volume of solid rotate the curve y=x+7 about y-axis X= 4-7 radius=x, aneo of dook =32c2 Volume of solid #44-78d4 = =[4-) = *[ Roze - 67-78] =*[**] = 90 Volume of resulting solid = 40.51-8T = 31.50 Volemo of required sold = 31.51) Content = 98.96 = X(4-7)2 (2) Calculation is more efficient when itegral is with respect to x by shell method. Since we get the volume by only one integral. (6) Yes to act up the integral with respect toy by dusk method the limit of regions are different so we calculate two integral to find the reparate volumes for each carne. In this case shell method will be preferred. Of region is rotated about x-axis then washer method will be preferred. So it depends upon the region to be rotated and axis about the region to be rotated.