Surface Area: y=sqrt(x) 3/4<=x<=15/4, around x-axis

[cheffy]

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!

 

[galactus]

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.

 

[soroban]

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.

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.