Area of a surface of revolution

[suspenc3]

9x=y^2+18 between 2&6..so y=(9x-18)^1/2

and dy/dx= $$\frac{9}{2 \sqrt{9x-18}}$$

so $$S = 2 \pi \int_2^6 \sqrt{9x-18} \sqrt{1+ \frac{81}{36x+72}}dx$$

If this is all right..then I am stuck

Any help?

 

[eigenglue]

Well first, you have a sign error in the denominator of the second term. Should be -72.

To integrate, you could mulitply the terms in the square roots and manipulate them algebraically to get something simpler.

 

[Gagle The Terrible]

$$S = 2 \pi \int_2^6 \sqrt{9x-18} \sqrt{1+ \frac{81}{36x-72}}dx$$

$$S = 6 \pi \int_2^6 \sqrt{x-1/4} }dx$$

Using the substitution u = x+1/4 you will get your answer

Hope it helps.

 

[pizzasky]

I think you should get $$S=6\pi\int_2^6\sqrt{x+\frac{1}{4}}dx$$

Also, since $$x+\frac{1}{4}$$ is linear, you can use the standard formula:-

$$\int(ax+b)^ndx=\frac{(ax+b)^{n+1}}{a(n+1)}+c$$, where n is unequal to -1.

 

[suspenc3]

$$S=6\pi\int_2^6\sqrt{x+\frac{1}{4}}dx$$

How did you get this?

 

[pizzasky]

Consider, $$\sqrt{9x-18} \sqrt{1+\frac{81}{36x-72}}=[(9x-18)(1+\frac{81}{36x-72})]^\frac{1}{2}$$

Carry out some factorization and re-expression and you should get what I got!