Difference between revisions of "009B Sample Final 1, Problem 5"

Latest revision as of 17:17, 2 December 2017

The region bounded by the parabola 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 y=x^2} and the line 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 y=2x} in the first quadrant is revolved about the 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 y} -axis to generate a solid.

(a) Sketch the region bounded by the given functions and find their points of intersection.

(b) Set up the integral for the volume of the solid.

(c) Find the volume of the solid by computing the integral.

Solution

Detailed Solution

Return to Sample Exam

@@ Line 7: / Line 7: @@
 <span class="exam">(c) Find the volume of the solid by computing the integral.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009B Sample Final 1, Problem 5 Solution|'''<u>Solution</u>''']]
-|-
-|'''1.''' You can find the intersection points of two functions, say &nbsp; <math style="vertical-align: -5px">f(x),g(x),</math>
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; by setting &nbsp;<math style="vertical-align: -5px">f(x)=g(x)</math>&nbsp; and solving for &nbsp;<math style="vertical-align: 0px">x.</math>
-|-
-|'''2.''' The volume of a solid obtained by rotating an area around the &nbsp;<math style="vertical-align: -4px">y</math>-axis using cylindrical shells is given by
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\int 2\pi rh~dx,</math>&nbsp; where &nbsp;<math style="vertical-align: 0px">r</math>&nbsp; is the radius of the shells and &nbsp;<math style="vertical-align: 0px">h</math>&nbsp; is the height of the shells.
-|}
-'''Solution:'''
+[[009B Sample Final 1, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']]
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|First, we sketch the region bounded by the given functions.
-|-
-|Insert graph here.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Setting the equations equal, we have &nbsp;<math style="vertical-align: 0px">x^2=2x.</math>
-|-
-|Solving for &nbsp;<math style="vertical-align: -4px">x,</math>&nbsp; we get
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
-\displaystyle{0} & = & \displaystyle{x^2-2x}\\
-&&\\
-& = & \displaystyle{x(x-2).}
-\end{array}</math>
-|-
-|So, &nbsp;<math style="vertical-align: 0px">x=0</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">x=2.</math>
-|-
-|If we plug these values into our functions, we get the intersection points
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -4px">(0,0)</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">(2,4).</math>
-|-
-|This intersection point can be seen in the graph shown in Step 1.
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|We proceed using cylindrical shells. The radius of the shells is given by &nbsp;<math style="vertical-align: 0px">r=x.</math>
-|-
-|The height of the shells is given by &nbsp;<math style="vertical-align: 0px">h=2x-x^2.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|So, the volume of the solid is
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -14px">\int 2\pi rh~dx~=~\int_0^2 2\pi x(2x-x^2)~dx.</math>
-|}
-'''(c)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|We need to integrate
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <math>\int_0^2 2\pi x(2x-x^2)~dx~=~2\pi\int_0^2 2x^2-x^3~dx.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|We have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
-\displaystyle{\int_0^2 2\pi x(2x-x^2)~dx} & = & \displaystyle{2\pi\int_0^2 2x^2-x^3~dx}\\
-&&\\
-& = & \displaystyle{2\pi\bigg(\frac{2x^3}{3}-\frac{x^4}{4}\bigg)\bigg|_0^2}\\
-&&\\
-& = & \displaystyle{2\pi\bigg(\frac{2^4}{3}-\frac{2^4}{4}\bigg)-2\pi(0)}\\
-&&\\
-& = & \displaystyle{\frac{8\pi}{3}.}\\
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp;'''(a)''' &nbsp; &nbsp;<math style="vertical-align: -5px">(0,0),(2,4)</math> (See Step 1 for the graph)
-|-
-|&nbsp; &nbsp;'''(b)''' &nbsp; &nbsp;<math style="vertical-align: -15px">\int_0^2 2\pi x(2x-x^2)~dx</math>
-|-
-|&nbsp; &nbsp;'''(c)''' &nbsp; &nbsp;<math style="vertical-align: -14px">\frac{8\pi}{3}</math>
-|}
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009B Sample Final 1, Problem 5"

Latest revision as of 17:17, 2 December 2017

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