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Calc II - CU Boulder - Fall 2010 - Section005

Book Assignment 6

Due Date: 
Monday, October 4, 2010 - 16:00

Section 8.4 Problems 10, 17
Section 9.1 Problems 44, 60
Section 9.2 Problem 30

Area and Arc Length in Polar Coordinates

Wednesday, September 22, 2010 - 14:00 - Friday, September 24, 2010 - 15:00

In this section, we introduced the formula for integrals in polar coordinates. We also did a bit with arc length in polar coordinates (which is really the same as arc length in parametric coordinates). Again, the webwork problems are the best resource here and what I used to motivate lectures in class.

Applications to Geometry

Monday, September 20, 2010 - 14:00 - Tuesday, September 21, 2010 - 15:00

This section was really an extension of 8.1. Some notes:

  • Remember that the integral's "dx" usually provides one of the dimensions (width, height, length) and so your integrand for area will be a single dimensional function, while your integrand for volume is a 2-dimensional function.
  • We did several examples about cross sections. The only way to get good at these problems is just to practice them. In any case, the last problem on webwork is a bit of a pain. You'll need the area of those triangles. This is easiest to do using the ideas from similar triangles and trigonometry. Specifically, remember that an equilateral triangle has three angles of 60 degrees, and sin(pi/3) is a nice number to work with.
  • We derived the arc length formulas. If you need those, you should derive or memorize those.

Book Assignment 5

Due Date: 
Monday, September 27, 2010 - 16:00

Section 8.2: Problems 44, 56
Section 8.3: Problems 32, 38

Midterm 1 Solutions

Math 2300 – Calculus II – University of Colorado

Fall 2010 – Midterm I – Solutions

  1. (26 points) Consider the integral \displaystyle\int _{0}^{{\sqrt{2}}}\sqrt{2-x^{2}}\, dx

    1. (6 points) Draw a picture of the area that this integral represents. Evaluate this integral any way you like.

      Solution: The graph of the function f(x)=\sqrt{2-x^{2}} for x in [0,\sqrt{2}] represents the upper right quarter (first quadrant) of the circle centered at the origin and of radius r=\sqrt{2}. We now want to compute the integral:

      Solution 1: Since the integrand f is positive on [0,\sqrt{2}], the integral \int _{0}^{{\sqrt{2}}}\sqrt{2-x^{2}}\, dx represents the area under the graph of f and above the x-axis; this is exactly the area of a quarter of the disc, which is \frac{1}{4}\pi r^{2}=\frac{2\pi}{4}=\frac{\pi}{2}.

      Solution 2: We make the trig substitution, and write x in [-\sqrt{2},\sqrt{2}] as x=\sqrt{2}\sin t, with t in [\frac{-\pi}{2},\frac{\pi}{2}]. Then the radical becomes: \sqrt{2-x^{2}}=\sqrt{2-2\sin^{2}t}=\sqrt{2\cos^{2}t}=\sqrt{2}\cos t. The integration limits for t are t=\sin^{{-1}}(0)=0 and t=\sin^{{-1}}(1)=\frac{\pi}{2}, and dx=\sqrt{2}\cos t\, dt. So:

      \displaystyle\int _{0}^{{\sqrt{2}}}\sqrt{2-x^{2}}\, dx \displaystyle=\int _{{0}}^{{\pi/2}}{\sqrt{2}\cos t\sqrt{2}\cos t\, dt}
      \displaystyle=\int _{{0}}^{{\pi/2}}{2\cos^{2}t\, dt}
      \displaystyle=\int _{{0}}^{{\pi/2}}{[1+\cos(2t)]\, dt}
      \displaystyle=\left.\left(t+\frac{\sin(2t)}{2}\right)\right\rvert _{0}^{{\pi/2}}
      \displaystyle\approx 1.57
    2. (6 points) Compute Left(2) and Mid(2). Show your work. An answer that appears to have come from a calculator will receive 0 points.


      \displaystyle\textsc{Left}(2) \displaystyle=\Delta x\left[f(x_{0})+f(x_{1})\right]
      \displaystyle\approx 1.86
      \displaystyle\textsc{Mid}(2) \displaystyle=\Delta x\left[f(\frac{x_{0}+x_{1}}{2})+f(\frac{x_{1}+x_{2}}{2})\right]
      \displaystyle\approx 1.63
    3. (4 points) The formula for error in an approximation is given by

      \textsc{Error }=\textsc{Actual Value }-\textsc{ Approximation}.

      Compute the errors in Left(2) and Mid(2).


      \textsc{Error Left}(2)\approx 1.57-1.86=-0.29
      \textsc{Error Mid}(2)\approx 1.57-1.62=-0.05
    4. (6 points) Estimate the errors in Left(20) and Mid(20) without actually computing these values. Again, an answer that appears to have come from a calculator will receive 0 points.


      \textsc{Error Left}(20)\approx-0.29/10=-0.029
      \textsc{Error Mid}(20)\approx-0.05/100=-0.0005
    5. (4 points) Order the following approximations from least to greatest for any integer n\ge 2.



      The function f is decreasing, so \textsc{Right}(n)<\displaystyle\int _{0}^{{\sqrt{2}}}\sqrt{2-x^{2}}\, dx<\textsc{Left}(n).

      The function f is concave down, so \textsc{Trap}(n)<\displaystyle\int _{0}^{{\sqrt{2}}}\sqrt{2-x^{2}}\, dx<\textsc{Mid}(n).

      Since Mid(n) and Trap(n) are generally closer estimates than Left(n) and Right(n), we can order:

      \textsc{Right}(n)<\textsc{Trap}(n)<\displaystyle\int _{0}^{{\sqrt{2}}}\sqrt{2-x^{2}}\, dx<\textsc{Mid}(n)<\textsc{Left}(n)

Areas and Volumes

Friday, September 17, 2010 - 14:00 - 15:00

I am posting WebWork (after it is due) solutions as the main resource for Sections 8.1-8.3. See the file attached to this lecture post.

One of the main ideas that we used in constructing the integrals and Riemann sums was that of similar triangles. If you are looking for more information about similar triangles, you might find the following link useful.


Worksheet 4

Solutions to worksheet 4 are posted.

Midterm 1 Solutions

Solutions to the first midterm are attached.

Midterm 1 Review #2

Math 2300 – Calculus II – University of Colorado

Fall 2010 – Review for Midterm I

This review is from instructor Anca Radulescu. I've included my own solutions up through the 'table of integrals' problems. (There may exist better, or more correct solutions.)

§ Substitution

Understand the concept of substitution (as a rewrite of the chain rule in terms of integrals), when to use it and how to carry it through (for both indefinite and definite integrals).

Some practice exercises: Using an appropriate substitution, show that the two integrals are identical:

  1. A\displaystyle{\int _{0}^{{1}}{f(Ax)dx}=\int _{0}^{{A}}{f(x)dx}}

    Solution: If w=Ax and dw=A\, dx then

    \displaystyle A\int _{0}^{1}f(Ax)\, dx \displaystyle=A\int _{{w=0}}^{{w=A}}f(w)\frac{dw}{A}
    \displaystyle=\int _{0}^{A}f(w)\, dw
    \displaystyle=\int _{0}^{A}f(x)\, dx

Gateway Exam Info

Thursday, September 23, 2010 - 14:00 - 15:00

The gateway exam will be held on September 23 during class. No calculators are allowed on this exam.

© 2011 Jason B. Hill. All Rights Reserved.