Arc lengh and surface area (integration)

[Brain0991]

Find the arc length of the graph of f(x)=(3/2)x^(2/3)+4 over [1,27]

Therefore:
dy/dx = 1/(x^(1/3))

? sqrt(1+(dy/dx)^2)dx
? sqrt(1+(x^(-1/3))^2)dx
? sqrt(1+1/(x^(2/3)))dx

And that is as far as I can go. I canot even follow the example in my book that is similar to this problem.




A sphere of radius r is generated by revolving the graph of y=sqrt(r^2-x^2) about the x-axis. Find the surface area of the solid generated.

hmm... there is an "r" still in the equation and I guess that just freaks me out, I do not even know where to begin.

 

[galactus]

Find the arc length of the graph of f(x)=(3/2)x^(2/3)+4 over [1,27]

Therefore:
dy/dx = 1/(x^(1/3))

? sqrt(1+(dy/dx)^2)dx
? sqrt(1+(x^(-1/3))^2)dx
? sqrt(1+1/(x^(2/3)))dx

And that is as far as I can go. I canot even follow the example in my book that is similar to this problem.

This is not as bad as it looks.

\(\displaystyle \sqrt{1+\frac{1}{x^{\frac{2}{3}}}}=\frac{\sqrt{1+x^{\frac{2}{3}}}}{x^{\frac{1}{3}}}\)


Make the substitution \(\displaystyle u=1+x^{\frac{2}{3}}, \;\ \frac{3}{2}du=\frac{1}{x^{\frac{1}{3}}}dx\)

Then, it whittles down to an easy integration.

A sphere of radius r is generated by revolving the graph of y=sqrt(r^2-x^2) about the x-axis. Find the surface area of the solid generated.

hmm... there is an "r" still in the equation and I guess that just freaks me out, I do not even know where to begin.

Don't be scared of the r. It's a constant. Plug a 1 in for r until you're done if it helps. Integrate w.r.t x all the same. Your result will be in terms of r. Yor should get the surface area of a sphere formula.