Difference between revisions of "Prototype Calculus Question"
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!Step 1: | !Step 1: | ||
|- | |- | ||
− | |'''Choosing the Approach:''' Since we are rotating around the ''x''-axis, the washer method would utilize tall rectangles with ''dx'' as their width. This seems like a reasonable choice, as these rectangles would be trapped between our two functions as ''x'' varies over the enclosed region, allowing us to solve a single integral. | + | |'''Choosing the Approach:''' Since we are rotating around the ''x''-axis, the washer method would utilize tall rectangles with ''dx'' as their width. This seems like a reasonable choice, as these rectangles would be trapped between our two functions as ''x'' varies over the enclosed region, allowing us to solve a single integral. Note that the washer method will require an inner and outer radius, as well as bounds of integration, in order to solve evaluate the integral |
+ | |- | ||
+ | |- | ||
+ | |<math>V=\pi \int_{x_1}^{\, x_2} R^2-r^2 \,dx</math> | ||
|} | |} | ||
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|<math>5-x=25-x^2,</math> | |<math>5-x=25-x^2,</math> | ||
|- | |- | ||
− | |then | + | |then by moving all terms to the left hand side and factoring, |
|- | |- | ||
− | |<math>x^2-x-20=(x+4)(x- | + | |<math>x^2-x-20=(x+4)(x-5)=0,</math> |
|- | |- | ||
− | |so we have | + | |so we have -4 and 5 as solutions. These are our bounds of integration. |
|} | |} | ||
Line 52: | Line 55: | ||
|'''Evaluating the Integral:''' Using the earlier steps, we have | |'''Evaluating the Integral:''' Using the earlier steps, we have | ||
|- | |- | ||
− | |<math>V=\pi \int_{-4}^{\, 5} (25-x^2)^2-(5-x)^2 \,dx</math> | + | |<math>V=\pi \int_{x_1}^{\, x_2} R^2-r^2 \,dx</math> |
+ | |- | ||
+ | | <math>=\pi \int_{-4}^{\, 5} (25-x^2)^2-(5-x)^2 \,dx</math> | ||
|- | |- | ||
| <math> =\pi \int_{-4}^{\, 5} 625-50x^2+x^4-(25-10x+x^2) \,dx</math> | | <math> =\pi \int_{-4}^{\, 5} 625-50x^2+x^4-(25-10x+x^2) \,dx</math> | ||
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| <math> =\pi \biggr(600x-51\cdot\frac{x^{3}}{3}+\frac{x^{5}}{5}+10\cdot\frac{x^{2}}{2}\biggr)\biggr|_{x=-4}^{5}</math> | | <math> =\pi \biggr(600x-51\cdot\frac{x^{3}}{3}+\frac{x^{5}}{5}+10\cdot\frac{x^{2}}{2}\biggr)\biggr|_{x=-4}^{5}</math> | ||
|- | |- | ||
− | | <math> =\frac{15,309}{5}\pi.</math> | + | | <math> =\frac{15,309}{5} \,\pi.</math> |
|} | |} |
Revision as of 20:36, 8 March 2015
Find the volume of the solid obtained by rotating the area enclosed by and
around the x-axis.
Foundations |
---|
• Choose either shell or washer method. |
• Find the appropriate radii. |
• Determine the bounds of integration by finding when both functions have the same y value. |
• Using the determined values, set up and solve the integral. |
Solution:
Step 1: |
---|
Choosing the Approach: Since we are rotating around the x-axis, the washer method would utilize tall rectangles with dx as their width. This seems like a reasonable choice, as these rectangles would be trapped between our two functions as x varies over the enclosed region, allowing us to solve a single integral. Note that the washer method will require an inner and outer radius, as well as bounds of integration, in order to solve evaluate the integral |
Step 2: |
---|
Finding the Radii: Since our rectangles will be trapped between the two functions, and will be rotated around the x-axis (where y = 0), we find |
the inner radius is , represented by the blue line, while |
the outer radius is , represented by the red line. |
Step 3: |
---|
Finding the Bounds of Integration: We must set the two functions equal, and solve. If |
then by moving all terms to the left hand side and factoring, |
so we have -4 and 5 as solutions. These are our bounds of integration. |
Step 4: |
---|
Evaluating the Integral: Using the earlier steps, we have |