{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Set5 - Math 21B-B Homework Set 5 Section 6.1 4 Lets...

Info icon This preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 21B-B - Homework Set 5 Section 6.1: 4. Let’s consider the area of a cross section of the solid when - 1 x 1. The diameter of the circular cross section at x is given by Diameter = ( 2 - x 2 ) - x 2 = 2 - 2 x 2 . Thus, the area of the cross section is given by Area = π · 1 2 ( 2 - 2 x 2 ) 2 = π · ( 1 - x 2 ) 2 = π · ( 1 - 2 x 2 + x 4 ) . Volume = Z 1 - 1 π · ( 1 - 2 x 2 + x 4 ) dx = π Z 1 - 1 ( 1 - 2 x 2 + x 4 ) dx = π · x - 2 3 x 3 + 1 5 x 5 1 - 1 = π · 1 - 2 3 + 1 5 - π · - 1 + 2 3 - 1 5 = π · 2 - 4 3 + 2 5 = 16 15 π 7. a. Let’s consider the area of a cross section of the solid when 0 x π . The base of the triangular cross section at x is given by Base = 2 sin x . The height of the triangle is given by Height = 3 · sin x . ( Use properties of equilateral triangles ) Therefore, the area of the cross section is given by Area = ( 1 2 ) 2 sin x 3 sin x = 3 sin x . 1
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Volume = Z π 0 3 sin x dx = - 3 cos x π 0 = 3 - ( - 3) = 2 3 b. Let’s consider the area of a cross section of the solid when 0 x π . The base of the square cross section at x is given by Base = 2 sin x . Therefore, the area of the cross section is given by Area = (2 sin x ) 2 = 4 sin x . Volume = Z π 0 4 sin x dx = - 4 cos x | π 0 = 4 - ( - 4) = 8 12. a. Diameter = 2 4 1 - x 2 Area = π · h 1 2 2 4 1 - x 2 i 2 = π 1 - x 2 Volume = Z 2 / 2 - 2 / 2 π 1 - x 2 dx = π Z 2 / 2 - 2 / 2 1 1 - x 2 dx = ( π · sin - 1 x ) 2 / 2 - 2 / 2 = π · π 4 - π · - π 4 = 2 · π 2 4 = π 2 2 2
Image of page 2
b. Diagonal = 2 4 1 - x 2 Area = r 1 4 1 - x 2 2 + 1 4 1 - x 2 2 ! 2 = 2 1 - x 2 Volume = Z 2 / 2 - 2 / 2 2 1 - x 2 dx = ( 2 sin - 1 x ) 2 / 2 - 2 / 2 = 2 · π 4 - 2 · - π 4 = 4 · π 4 = π 15. We are rotating about the x -axis, so we want to take the integral with respect to x . We want 0 x 2. The radius of a cross section is given by Radius = - 1 2 x + 1 ( Between graph and x -axis ) Therefore, the area of the cross section is given by Area = π · ( - 1 2 x + 1 ) 2 = π · ( 1 4 x 2 - x + 1 ) Volume = Z 2 0 π · 1 4 x 2 - x + 1 dx = π · 1 12 x 3 - 1 2 x 2 + x 2 0 = π · 8 12 - 2 + 2 = 2 π 3 16. We are rotating about the y -axis, so we want to take the integral with respect to y . We want 0 y 2. The radius of the cross section is given by Radius = 3 y 2 ( Between the graph and the y -axis ) Therefore, the area of the cross section is given by 3
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Area = π · ( 3 y 2 ) 2 = π · 9 y 2 4 Volume = Z 2 0 π · 9 y 2 4 dy = π · 3 y 3 4 2 0 = 6 π 20. We are rotating about the x -axis, so we want to take the integral with respect to y . We want 0 x 2. The radius of the cross section is given by Radius = x 3 Therefore, the area of the cross section is given by Area = π ( x 3 ) 2 = π · x 6 Volume = Z 2 0 π · x 6 dx = π 7 x 7 2 0 = 128 π 7 29. We are rotating about the line y = 2, so we want to take the integral with respect to x . We want 0 x 2. The radius of the cross section is given by Radius = 2 - sec x tan x ( Between y = 2 and y = sec x tan x ) Area = π · ( 2 - sec x tan x ) 2 = π · ( 2 - 2 2 sec x tan x + sec 2 x tan 2 x ) 4
Image of page 4
Volume = Z π/ 4 0 π · 2 - 2 2 sec x tan x + sec 2 x tan 2 x dx = π · 2 x - 2 2 sec x + 1 3 tan 3 x π/ 4 0 ( Use u-substitution ) = π · π 2 - 2 2 · 2 + 1 3 - π · - 2 2
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern