What are you getting and how are you getting it?

Domain is [-3,1]

Diameter is

Radius is

Area of Entire Circle is

Area of Semi-Circle is

Results 1 to 14 of 14

- July 12th 2008, 01:51 PM #1

- Joined
- Jul 2008
- Posts
- 15

## 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.

- July 12th 2008, 02:12 PM #2

- Joined
- Aug 2007
- From
- USA
- Posts
- 3,111
- Thanks
- 2

- July 12th 2008, 02:14 PM #3

- Joined
- Jul 2008
- Posts
- 15

- July 12th 2008, 02:16 PM #4
Did you draw the region? We need this to help us to determine the value for our radius [so we can find its volume]

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

We see that it has a width of .

Thus the radius of each circle would be

As a result, each semi circle will have an area of

We also need to know the points of intersection:

Thus, the volume of our solid would be

Can you take it from here? You will get as your answer.

Hope that this makes sense!

--Chris

- July 12th 2008, 02:19 PM #5

- Joined
- Jul 2008
- Posts
- 15

- July 12th 2008, 02:21 PM #6

- Joined
- Aug 2007
- From
- USA
- Posts
- 3,111
- Thanks
- 2

- July 12th 2008, 02:54 PM #7

- Joined
- Jul 2008
- Posts
- 15

- July 12th 2008, 04:01 PM #8

- Joined
- Aug 2007
- From
- USA
- Posts
- 3,111
- Thanks
- 2

I understand everything except I keep getting the wrong answer.

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?

SHOW YOUR WORK.

- July 12th 2008, 05:51 PM #9

- Joined
- Jul 2008
- Posts
- 15

- July 12th 2008, 06:00 PM #10

- July 12th 2008, 06:02 PM #11

- July 12th 2008, 06:09 PM #12

- Joined
- Jul 2008
- Posts
- 15

- July 12th 2008, 06:13 PM #13

- July 12th 2008, 07:33 PM #14

- Joined
- Aug 2007
- From
- USA
- Posts
- 3,111
- Thanks
- 2

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.