C cheffy Junior Member Joined Jan 10, 2007 Messages 73 Mar 2, 2007 #1 Find the surface area: y=sqrt(x) 3/4<=x<=15/4 rotated around the x-axis I made r=y=sqrt(x) and ds=sqrt((4x+1)/(4x))dx Does this make sense? I'm getting weird integrals for dS. Thanks!
Find the surface area: y=sqrt(x) 3/4<=x<=15/4 rotated around the x-axis I made r=y=sqrt(x) and ds=sqrt((4x+1)/(4x))dx Does this make sense? I'm getting weird integrals for dS. Thanks!
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Mar 2, 2007 #2 What you did looks OK. \(\displaystyle \L\\y'=\frac{d}{dx}[\sqrt{x}]=\frac{1}{2\sqrt{x}}\) \(\displaystyle \L\\2{\pi}\int_{\frac{3}{4}}^{\frac{15}{4}}\sqrt{x}\sqrt{1+\frac{1}{4x}}dx\) \(\displaystyle =\L\\2{\pi}\int_{\frac{3}{4}}^{\frac{15}{4}}\frac{\sqrt{4x+1}}{2}dx\) Let \(\displaystyle u=4x+1, \;\ du=4dx, \;\ \frac{du}{4}=dx\) \(\displaystyle \L\\2{\pi}\int_{4}^{16}\frac{\sqrt{u}}{8}du\) Now integrate.
What you did looks OK. \(\displaystyle \L\\y'=\frac{d}{dx}[\sqrt{x}]=\frac{1}{2\sqrt{x}}\) \(\displaystyle \L\\2{\pi}\int_{\frac{3}{4}}^{\frac{15}{4}}\sqrt{x}\sqrt{1+\frac{1}{4x}}dx\) \(\displaystyle =\L\\2{\pi}\int_{\frac{3}{4}}^{\frac{15}{4}}\frac{\sqrt{4x+1}}{2}dx\) Let \(\displaystyle u=4x+1, \;\ du=4dx, \;\ \frac{du}{4}=dx\) \(\displaystyle \L\\2{\pi}\int_{4}^{16}\frac{\sqrt{u}}{8}du\) Now integrate.
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Mar 2, 2007 #3 Re: Surface Area Hello, cheffy! Find the surface area: \(\displaystyle \,y\:=\:\sqrt{x},\;\;\frac{3}{4}\,\leq\,x\,\leq\,\frac{15}{4}\), rotated around the x-axis. Click to expand... Your work is correct . . . \(\displaystyle y\:=\:\sqrt{x},\;ds\,=\,\sqrt{\frac{4x\,+\,1}{4x}} \:=\:\frac{\sqrt{4x\,+\,1}}{2\sqrt{x}}\) Then: \(\displaystyle \:S\;=\;2\pi\L\int^{\;\;\frac{15}{4}}_{\frac{3}{4}}\sqrt{x}\cdot\frac{\sqrt{4x\,+\,1}}{2\sqrt{x}}\)\(\displaystyle \,dx\;=\;\pi\L\int^{\;\;\frac{15}{4}}_{\frac{3}{4}}\)\(\displaystyle \sqrt{4x\,+\,1}\,dx\) Now let: \(\displaystyle u\:=\:4x\,+\,1\) . . . etc.
Re: Surface Area Hello, cheffy! Find the surface area: \(\displaystyle \,y\:=\:\sqrt{x},\;\;\frac{3}{4}\,\leq\,x\,\leq\,\frac{15}{4}\), rotated around the x-axis. Click to expand... Your work is correct . . . \(\displaystyle y\:=\:\sqrt{x},\;ds\,=\,\sqrt{\frac{4x\,+\,1}{4x}} \:=\:\frac{\sqrt{4x\,+\,1}}{2\sqrt{x}}\) Then: \(\displaystyle \:S\;=\;2\pi\L\int^{\;\;\frac{15}{4}}_{\frac{3}{4}}\sqrt{x}\cdot\frac{\sqrt{4x\,+\,1}}{2\sqrt{x}}\)\(\displaystyle \,dx\;=\;\pi\L\int^{\;\;\frac{15}{4}}_{\frac{3}{4}}\)\(\displaystyle \sqrt{4x\,+\,1}\,dx\) Now let: \(\displaystyle u\:=\:4x\,+\,1\) . . . etc.
