# I think that I am overthinking it

• Jul 12th 2008, 01:51 PM
jazz836062
I think that I am overthinking it
it is straight out of my book:

Find the volume of the solid where the baseis the given region and which has the property that each cross section perpendicular to the x-axis is a semicircle.

The region bounded by the parabola y = x^2 and the line 2x + y - 3 = 0

The answer is (64/15)pi but i don't know how to get it.
• Jul 12th 2008, 02:12 PM
TKHunny
What are you getting and how are you getting it?

Domain is [-3,1]

Diameter is $((3-2x)-x^{2})$

Radius is $\frac{((3-2x)-x^{2})}{2}$

Area of Entire Circle is $\pi*\left(\frac{((3-2x)-x^{2})}{2}\right)^{2}$

Area of Semi-Circle is $\frac{1}{2}*\pi*\left(\frac{((3-2x)-x^{2})}{2}\right)^{2}$
• Jul 12th 2008, 02:14 PM
jazz836062
That is how I have been working it. I have been getting. (544/15)pi I think.
• Jul 12th 2008, 02:16 PM
Chris L T521
Quote:

Originally Posted by jazz836062
it is straight out of my book:

Find the volume of the solid where the baseis the given region and which has the property that each cross section perpendicular to the x-axis is a semicircle.

The region bounded by the parabola y = x^2 and the line 2x + y - 3 = 0

The answer is (64/15)pi but i don't know how to get it.

Did you draw the region? We need this to help us to determine the value for our radius [so we can find its volume]

http://img.photobucket.com/albums/v4...LT521/Prob.jpg

The cyan colored part is the region we're dealing with.

We see that it has a width of $3-2x-x^2$.

Thus the radius of each circle would be $\frac{3-2x-x^2}{2}$

As a result, each semi circle will have an area of $\frac{\pi}{2}\cdot\bigg[\frac{3-2x-x^2}{2}\bigg]^2$

We also need to know the points of intersection:

$3-2x=x^2\implies x^2+2x-3=0\implies (x+3)(x-1)=0\implies x=-3 \ or \ x=1$

Thus, the volume of our solid would be $\frac{\pi}{8}\int_{-3}^1\bigg(3-2x-x^2\bigg)^2\,dx$

Can you take it from here? You will get $\frac{64\pi}{15}$ as your answer.

Hope that this makes sense! :D

--Chris
• Jul 12th 2008, 02:19 PM
jazz836062
Thank you. It makes perfect sense.
• Jul 12th 2008, 02:21 PM
TKHunny
Please observe, Jazz, that you showed no work during this conversation. Don't do that. SHOW YOUR WORK! Trust me on this.
• Jul 12th 2008, 02:54 PM
jazz836062
I understand everything except I keep getting the wrong answer. I keep getting $\pi*2011/240$. Could you walk me through your integration steps?
• Jul 12th 2008, 04:01 PM
TKHunny
Quote:

I understand everything except I keep getting the wrong answer.
You've got to see how silly that sounds. There is something you don't understand. Too bad no one knows what it is since you are showing no work.

Did you set up the integral?

Did you get the same integrand?

Did you pull out the constants or leave them inside the integral?

Did you square the trinomial or try to think of some other way to find the antiderivative?

Did you divide the integration into pieces or try to do it all in one piece?

• Jul 12th 2008, 05:51 PM
jazz836062
I expanded it out and got:

$\frac{\pi}{8}\int_{-3}^1\bigg(9 - 12x-2x^2 + 6x^3 + x^4\bigg), dx$

Then I integrated each part:
$(9x-6x^2-\frac{2x^3}{3} + \frac{3x^4}{2} + \frac{x^5}{5}$

That gives me $\frac{-88}{15}$
• Jul 12th 2008, 06:00 PM
Chris L T521
Quote:

Originally Posted by jazz836062
I expanded it out and got:

$9x-6x^2-\frac{2x^3}{3} + \frac{3x^4}{2} + \frac{x^5}{5}$

Its very close to what I have...I get $\int_{-3}^1 \bigg(3-2x-x^2\bigg)^2\,dx= \left.\left[9x-6x^2-\frac{2}{3}x^3+{\color{red}x^4}+\frac{1}{5}x^5\rig ht]\right|_{-3}^1$

Can you try to figure out where you went wrong? Showing your steps may be helpful.

--Chris

w00t!!! my 3(Sun)(Sun) post! :D
• Jul 12th 2008, 06:02 PM
Chris L T521
Quote:

Originally Posted by jazz836062
I expanded it out and got:

$\frac{\pi}{8}\int_{-3}^1\bigg(9 - 12x-2x^2 + {\color{red}6}x^3 + x^4\bigg), dx$

Then I integrated each part:
$\frac{\pi}{8}\int_{-3}^1\bigg(9x-6x^2-\frac{2x^3}{3} + \frac{3x^4}{2} + \frac{x^5}{5}\bigg), dx$

That gives me $\frac{-88}{15}$

It should be ${\color{red}4}x^3$...

--Chris
• Jul 12th 2008, 06:09 PM
jazz836062
$(3-2x-x^2)(3-2x-x^2)$

$9-6x-3x^2-6x+4x^2+2x^3-3x^2+2x^3+x^4$

$9-12x-2x^2+4x^3+x^4$

I am new to this kind of coding so I am very slow at posting.
• Jul 12th 2008, 06:13 PM
Chris L T521
Quote:

Originally Posted by jazz836062
$(3-2x-x^2)(3-2x-x^2)$

$9-6x-3x^2-6x+4x^2+2x^3-3x^2+2x^3+x^4$

$9-12x-2x^2+4x^3+x^4$

I am new to this kind of coding so I am very slow at posting.

Correct. Now integrate to get $9x-6x^2-\frac{2}{3}x^3+x^4+\frac{1}{5}x^5$. Now apply the FTC to evaluate the integral.

You should get $\frac{512}{15}$. Then multiply by $\frac{\pi}{8}$ to get the answer you're looking for.

Hope this helps.

--Chris
• Jul 12th 2008, 07:33 PM
TKHunny
I am delighted at this excellent example of the value of showing one's work.

When you get to trigonometric substitutions, this might be a nice challenge problem to attempt WITHOUT multiplying out the trinomial. It's not pretty, but might be an interesting exploration.