Once they understood that the thickness of the paper was dx, it was very easy to set up the integrals to calculate the volumes of their models. Here they are, taping the cross-sections onto the base area: Here are my students in action, cutting out the cross-sections: Students use the templates to cut out a cross-section that fits down the middle of the base area, and six others on each side. I had my students cut the square templates diagonally for isosceles right triangles, and horizontally for rectangles. In her session, Nina Otterson provided templates that fit the given base area for different shapes: semicircles, squares, and equilateral triangles. Here’s what the base area looks like, courtesy of ’s online function grapher: She has her students cut cross-sections of different shapes and apply them to a base area enclosed by two parabolas, y = x^2 – 3 and y = 3 – x^2. Nina Chung Otterson was the presenter, and she teaches at The Hotchkiss School in Connecticut. Last year, I went to the regional NCTM conference here in Nashville, TN, and one of the sessions I attended addressed this exact issue. It’s hard, because they have difficulty visualizing it. Solving this equation for y, we see that y = sqrt( 9 - 9/16*x 2 )īut, if we graph out the ellipse, we can see that the coordinate "y" is only half of the total length of the hypotenouse.One of the hardest type of problem for calculus students to understand is calculating the volume of solids of known cross-sections. V = integral of ( 1/4 * d ^ 2 ) from (x=-4) to (x=4) dxīut we still need to express the length of the hypotenuse of our triangle, d, in terms of the parameter that we are integrating, in this case that parameter is x. The bounds of our integral x1 and x2 we can find by just plugging in y = 0 to our elliptical formula and getting x=-4 and x=4. V = integral of ( 1/4 * d ^ 2 ) from (x=x1) to (x=x2) dx Because our triangles lie perpendicular to the x-axis, we need to integrate the area along the x-axis to find the volume. The volume of any solid with a cross section of an isosceles right triangle is just the integral of that area along the length of that solid. The area of an Isosceles right triangle is, when given the length of the diagonal, d, just equal to Pastebin / / IDEOne Revert to older template If you have trouble with spacing, add some blank spaces with įor more commands, please see User Moderation Please contact u/Oryv if the bot malfunctions, or if you have questions on how to use it. Simply mention the bot in a comment and put the expressions that need to be rendered in code blocks, and u/LaTeX4Reddit will respond. To use LaTeX on this subreddit, use u/LaTeX4Reddit. Accept answers at your own risk.įor citation questions, check the Purdue Online Writing Lab Keep in mind that we do not and will not have any sort of vetting procedure for responders. This includes asking for "likes," page views, or similar things. Offers or solicitations of payment in any form. Requests for help with cheating, plagiarism, or other violations of academic integrity violations of copyright or terms of use or other illegal or unethical activities. Posts tagged "urgent," "ASAP," "important," "due in an hour," etc. "Do this for me" posts such as posting of quizzes or lists of questions. When your question has been answered, please edit the post's flair to " ✓ Answered".Ĭopied questions without context or explanation. What does your instructor want you to accomplish? Still acceptable…, but preferably not: Grade 1-6 (Pri) Grade 7-10 (Sec) Grade 11-12 (Pre-Uni) University/College ❗️ READ THE RULES BEFORE POSTING
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