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- Mar 18th 2006, 11:54 AMfrozenflamesCalculus Problem
Please See Attached

- Mar 18th 2006, 03:57 PMThePerfectHacker
You are going to be using the formula

$\displaystyle V=\int^b_aA(x)dx$

First see that you are taking the integral from $\displaystyle x=0$ until $\displaystyle x=8$ that is when it intersects the x and y axes.

Now you need to find $\displaystyle A(x)$:

Notice that $\displaystyle x+2y=8$ implies $\displaystyle y=-\frac{1}{2}x+8$.

At $\displaystyle x$ the width of the line perpendicular to the x-axis is then $\displaystyle -\frac{1}{2}x+8$. But this is the diameter of the circle. Thus, its radius is $\displaystyle -\frac{1}{4}x+4$.

The area of this cross section (semicircle) is

$\displaystyle \pi \frac{1}{2}\left(-\frac{1}{4}x+4\right)^2$ (remember the 1/2 it is semi-circle :eek: )

Thus,

$\displaystyle V=\pi \int^8_0\frac{1}{2}\left(-\frac{1}{4}x+4\right)^2dx=\pi \cdot 37.\bar 3\approx 116$

I think there is something wrong with your choices. - Mar 18th 2006, 05:42 PMRich B.
Dear PHckr:

Line #5 of your narrative reads:

Notice that http://www.mathhelpforum.com/math-he...db6e2cfd-1.gif implies http://www.mathhelpforum.com/math-he...428afcca-1.gif.

Do you see the problem?

Regards,

Rich B.

- Mar 19th 2006, 10:24 AMThePerfectHackerQuote:

Originally Posted by**Rich B.**