Solid of Revolution, Arc Length and Surface of Revolution

[Pre]

I got this worksheet and I'm having trouble. . .one problem is find the length of y=square root(3x-2) from x=1 to x=6. I know the formula is the integral from a to b of the square root of (1 + derivitive squared). I know the derivitive is 3/2(3x-2)^-.5

But I'm just making stupid mistakes because I cant get the right answer.

 

[skeeter]

\(\displaystyle \L y = \sqrt{3x-2}\)

\(\displaystyle \L y' = \frac{3}{2\sqrt{3x-2}}\)

\(\displaystyle \L (y')^2 = \frac{9}{4(3x-2)}\)

\(\displaystyle \L 1 + (y')^2 = \frac{4(3x-2)}{4(3x-2)} + \frac{9}{4(3x-2)} = \frac{12x+1}{12x-8}\)

\(\displaystyle \L S = \int_1^6 \sqrt{\frac{12x+1}{12x-8}} dx \approx 5.907\)

 

[Pre]

Thanks a lot! There is another problem I need help on.

The base of a solid is bounded by y=x^2 and y=2x+3. Find the volume of the solid if every cross section perpendicular to the major axis is an equilateral triangle.

I know the area of the triangle would be sqrt(3)/4 times base^2. Would the base be upper - lower, so [(2x+3)-(x^2)]^2 ?[/tex]

 

[skeeter]

that'll work if the "major" axis is the x-axis.