1. ## Confusing Basic Integration

The diagram shows part of the curves '$\displaystyle y^2=8x$' and $\displaystyle y=x^2$

a.) Find the coordinates of A ( A is where they intercept in the top right of the diagram)

b.) Calculate the volume generated when the area enclosed by the curve is rotated through 360(degrees) about the x axis.

I identified A to be (2,4).

Then to attempt the 'b' part of my question

$\displaystyle \pi$$\displaystyle $\int_0^2 [(x^2)-sqrt(8x))]^2 \,dx.$$

Using this I get a result of $\displaystyle 8\pi/2$ which is incorrect. Wondering if you can see my error. Sorry for the lack of steps, I'm having difficulty with LATEX (I'm in the process of improving though).

2. Since you are squaring the integrand the order does not matter but it is best to think of this as

$\displaystyle \displaystyle \pi\int_{0}^{2}[\sqrt{8x}-x^2]^2dx=\frac{144\pi}{35}$

As you said since you didn't show your steps I don't know where the error occurred.

3. Originally Posted by TheEmptySet
Since you are squaring the integrand the order does not matter but it is best to think of this as

$\displaystyle \displaystyle \pi\int_{0}^{2}[\sqrt{8x}-x^2]^2dx=\frac{144\pi}{35}$

As you said since you didn't show your steps I don't know where the error occurred.
Is this set up correct from the data given in the question?

Okay thanks for your help, however my textbook has that result down as wrong too.

4. No I am incorrect!!

It should be

$\displaystyle \displaystyle \int_{0}^{2}[\sqrt{8x}]^2-[x^2]^2dx=\frac{48}{5}\pi$

Sorry