
New Member
Volume of a solid (integration word problem)
Find the volume of a solid whose base is enclosed by the circle x^2+y^2 = 9 and whose cross sections taken perpendicular to the xaxis are as follows:
a. Semicircles
b. Equilateral triangles
Solution
ill just let the equation of circle be in terms of x since the area sweep is parallel to the x axis
a.
Since the lower limit of y is zero then the radius will just be
[TEX]y = \sqrt{9x^2}[/TEX]
Area of semicircle
[TEX]A = \frac{1}{2}\pi{r}^2[/TEX]
then Volume
[TEX]V = \int_{3}^{3}{A}dx[/TEX]
[TEX]V = \int_{3}^{3}{\frac{1}{2}\pi{r}^2}dx[/TEX]
[TEX]V = \int_{3}^{3}{\frac{1}{2}\pi(\sqrt{9x^2})^2}dx[/TEX]
[TEX]V = 18\pi[/TEX]
b.
The lower portion and upper portion is
[TEX]y = \sqrt{9x^2}  (\sqrt{9x^2})[/TEX]
[TEX]s=y = 2\sqrt{9x^2}[/TEX]
Area of equilateral
[TEX]A = \frac{(s)^2\sqrt{3}}{4}[/TEX]
Volume
[TEX]V = \int_{3}^{3}{A}dx[/TEX]
[TEX]V = \int_{3}^{3}{\frac{(2\sqrt{9x^2})^2\sqrt{3}}{4}}dx[/TEX]
[TEX]V = 62.35[/TEX]
i know this is the wrong answer because its not in the multiple choice answers
1.)
a. 66pi
b. 70pi
c. 68pi
d. 72pi
2.)
a. 16sqrt(3)
b. 17sqrt(3)
c. 18sqrt(3)
d. 19sqrt(3)
Last edited by ZyzzBrah; 12212011 at 06:59 AM.

Elite Member
Part a, I do not think you're wrong. I get [tex]18\pi[/tex] as well.
Your work and logic seems OK. Same with part b.
The choices given appear to be multiples of your answers.

Elite Member
Estimate with a sphere of radius 3.
[TEX]\frac{1}{2}\cdot\frac{4}{3}\pi 3^{3} = 18\pi[/TEX]
Okay, was that really an "estimate"?
Really, ALWAYS find a way to verify your work. In your "a", substitite +/ 3 with +/ r and substitute 9 with r^2 and see what you produce. It should look familiar.
Last edited by tkhunny; 12212011 at 02:55 PM.
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