Volume Problems

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• Nov 28th 2010, 01:40 PM
wair
Volume Problems
a). find volume of pie with height 2 inches, a base radius of 4.5 inches and a top radius of 5.5 inches

b). Find the approximate volume of a football given the following dimensions. The regulation football is a prolated spheroid with a long axis 11 inches and a circular circumference at the center of the ball of 21 inches. let us make the assumption that when looking at the football from our side, the top half looks like a parabola. write an equation for this parabola and use it to help find the volume of the football.

For part a what I did was add the two circle areas together and multiply with the height please correct me if I am wrong.

For part b I think the equation is y= x^2 -11* 2pi(21). I know part b is wrong, so if you could help me it would be appreciated. thank you.
• Nov 28th 2010, 01:43 PM
dwsmith
Part a: $\displaystyle V=\frac{\pi(r^2+rR+R^2)h}{3}$
• Nov 28th 2010, 02:16 PM
wair
Thank you dwsmith. emm could you explain how you got that ? if you don't mind. Also any idea how to solf part b ?
• Nov 28th 2010, 02:30 PM
dwsmith
That is the formula for a Frustum of Right Circular Cone. I just looked it up since that is the shape you described.

For b, I would put the parabola on the x axis where x intercept is $\pm 5.5$ since the $C=21=2\pi r\rightarrow r=\frac{21}{2\pi}$

r would be are y intercept. Now you have your intercepts. Find the equation of the parabola now.

Or you can have the x intercepts be 0 and 11 and y is 0 as well. Might be easier.
• Nov 28th 2010, 02:46 PM
wair
Frustum of right circular cone ? I always thought pies were circles. But thank you. And thanks for part b really helped.
• Nov 28th 2010, 02:59 PM
wair
Don't I need the axis of symmetry in order to write the equation of a parabola ?
• Nov 28th 2010, 03:03 PM
dwsmith
If you have the x and y intercepts, you should be able to solve it or -a(x-h)^2+k (h,k) is the vertex. You can use the second method if you want since we know the vertex is at (0, 21/(2pi)) and it crosses the x axis at plus and minus 5.5
• Nov 28th 2010, 03:11 PM
wair
Hmm I tried solving it with only the intercepts, I get y= -x^2 + 32.99. However that way the x intercepts are + or - 5.7. Is this correct ?
• Nov 28th 2010, 03:19 PM
dwsmith
$\displaytyle a(x-h)^2+k=y$

$\displaytyle a(x)^2+\frac{21}{2\pi}=y$

When y=0, x=-5.5 and 5.5

$\displaystyle x=\sqrt{\frac{21}{2a\pi}}=\pm 5.5$ Solve for a.
• Nov 28th 2010, 03:35 PM
wair
I still Don't understand how would you figure out the equation through intercepts alone ? When I solve for a I get 4.36 I'm guessing I plug that in to get why then plug y in to get h correct ?
• Nov 28th 2010, 03:37 PM
wair
By randomly entering values into my calculator I get y= -1.09x^2 + 21/2(pi), This equation satisfies all the intercepts,but i'd still like to know how to properly find this equation without guessing.
• Nov 28th 2010, 03:41 PM
dwsmith
When $\displaystyle x=\pm 5.5$ y must equal 0.

General parabola form:

$\displaysyle a(x-h)^2+k=y$

$\displaystyle(h,k)= \left(0,\frac{21}{2\pi}\right)$

$\displaystyle ax^2+\frac{21}{2\pi}=y\rightarrow a(5.5)^2+\frac{21}{2\pi}=0\rightarrow a=\frac{-21}{2*5.5^2\pi}$

$\displaystyle \frac{-21}{2*5.5^2\pi}x^2+\frac{21}{2\pi}=y$
• Nov 28th 2010, 04:09 PM
wair
When I graph that the x intercept isn't 5.5. This is how I am putting it into my calculator: y= -0.1104877x^2+32.9867
• Nov 28th 2010, 04:10 PM
dwsmith
Put in the fractions not the decimal approximation and it works.
• Nov 28th 2010, 04:11 PM
dwsmith
Also, the y intercept is 3.34225380 not in the 30s

It is $\displaystyle \frac{21}{2\pi} \ \mbox{not} \ \frac{21}{2}*\pi$
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