Difference between revisions of "009B Sample Final 3, Problem 4"

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<span class="exam"> Find the volume of the solid obtained by rotating about the &nbsp;<math>x</math>-axis the region bounded by &nbsp;<math style="vertical-align: -4px">y=\sqrt{1-x^2}</math>&nbsp; and &nbsp;<math>y=0.</math>
+
<span class="exam"> Find the volume of the solid obtained by rotating about the &nbsp;<math>x</math>-axis the region bounded by &nbsp;<math style="vertical-align: -4px">y=\sqrt{1-x^2}</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">y=0.</math>
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
|We start by finding the intersection points of the functions <math>y=\sqrt{1-x^2}</math> and <math>y=0.</math>
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|We start by finding the intersection points of the functions &nbsp;<math style="vertical-align: -4px">y=\sqrt{1-x^2}</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">y=0.</math>
 
|-
 
|-
|We need to solve <math>0=\sqrt{1-x^2}.</math>
+
|We need to solve  
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math>0=\sqrt{1-x^2}.</math>
 
|-
 
|-
 
|If we square both sides, we get  
 
|If we square both sides, we get  
 
|-
 
|-
|<math>0=1-x^2.</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>0=1-x^2.</math>
 
|-
 
|-
|The solutions to this equation are <math>x=-1</math> and <math>x=1.</math>
+
|The solutions to this equation are &nbsp;<math style="vertical-align: -1px">x=-1</math>&nbsp; and &nbsp;<math style="vertical-align: -1px">x=1.</math>
 
|-
 
|-
|Hence, we are interested in the region between <math>x=-1</math> and <math>x=1.</math>
+
|Hence, we are interested in the region between &nbsp;<math style="vertical-align: -1px">x=-1</math>&nbsp; and &nbsp;<math style="vertical-align: -1px">x=1.</math>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|Using the disk method, the radius of each disk is given by <math>r=\sqrt{1-x^2}.</math>
+
|Using the disk method, the radius of each disk is given by &nbsp;<math style="vertical-align: -3px">r=\sqrt{1-x^2}.</math>
 
|-
 
|-
 
|Therefore, the volume of the solid is
 
|Therefore, the volume of the solid is

Revision as of 11:29, 3 March 2017

Find the volume of the solid obtained by rotating about the  -axis the region bounded by    and  

Foundations:  
1. You can find the intersection points of two functions, say  

        by setting    and solving for  

2. The volume of a solid obtained by rotating a region around the  -axis using disk method is given by

          where    is the radius of the disk.


Solution:

Step 1:  
We start by finding the intersection points of the functions    and  
We need to solve
       Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 0={\sqrt {1-x^{2}}}.}
If we square both sides, we get
       
The solutions to this equation are  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=-1}   and  
Hence, we are interested in the region between  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=-1}   and  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=1.}
Step 2:  
Using the disk method, the radius of each disk is given by  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r=\sqrt{1-x^2}.}
Therefore, the volume of the solid is
        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{V} & = & \displaystyle{\int_{-1}^1 \pi (\sqrt{1-x^2})^2~dx}\\ &&\\ & = & \displaystyle{\int_{-1}^1 \pi (1-x^2)~dx}\\ &&\\ & = & \displaystyle{\pi\bigg(x-\frac{x^3}{3}\bigg)\bigg|_{-1}^1}\\ &&\\ & = & \displaystyle{\pi\bigg(1-\frac{1}{3}\bigg)-\pi\bigg(-1+\frac{1}{3}\bigg)}\\ &&\\ & = & \displaystyle{\frac{4\pi}{3}.} \end{array}}


Final Answer:  
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{4\pi}{3}}

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