Difference between revisions of "Prototype Calculus Question"

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!Step 1:  
 
!Step 1:  
 
|-
 
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|'''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 evaluate the integral
 +
|-
 +
|-
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|&nbsp;&nbsp;<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-20)=0,</math>
+
|<math>x^2-x-20=(x+4)(x-5)=0,</math>
 
|-
 
|-
|so we have roots -4 and 5.  These are our bounds of integration.
+
|so we have -4 and 5 as solutions.  These are our bounds of integration.
 
|}
 
|}
  
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|'''Evaluating the Integral:''' &nbsp; Using the earlier steps, we have
 
|'''Evaluating the Integral:''' &nbsp; 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>
 +
|-
 +
|&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<math>=\pi \int_{-4}^{\, 5} (25-x^2)^2-(5-x)^2 \,dx</math>
 
|-
 
|-
 
|&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<math> =\pi \int_{-4}^{\, 5} 625-50x^2+x^4-(25-10x+x^2) \,dx</math>
 
|&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<math> =\pi \int_{-4}^{\, 5} 625-50x^2+x^4-(25-10x+x^2) \,dx</math>
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|&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<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>
 
|&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<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>
 
|-
 
|-
|&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<math> =\frac{15,309}{5}\pi.</math>
+
|&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<math> =\frac{15,309}{5} \,\pi.</math>
 
|}
 
|}

Latest revision as of 19:30, 10 March 2015

9BSF1 3a.png

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 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