# Thread: Volume of a solid (integration word problem)

1. ## 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 x-axis 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
$y = \sqrt{9-x^2}$

Area of semicircle
$A = \frac{1}{2}\pi{r}^2$
then Volume
$V = \int_{-3}^{3}{A}dx$
$V = \int_{-3}^{3}{\frac{1}{2}\pi{r}^2}dx$
$V = \int_{-3}^{3}{\frac{1}{2}\pi(\sqrt{9-x^2})^2}dx$
$V = 18\pi$
b.

The lower portion and upper portion is
$y = \sqrt{9-x^2} - (-\sqrt{9-x^2})$
$s=y = 2\sqrt{9-x^2}$
Area of equilateral
$A = \frac{(s)^2\sqrt{3}}{4}$
Volume
$V = \int_{-3}^{3}{A}dx$
$V = \int_{-3}^{3}{\frac{(2\sqrt{9-x^2})^2\sqrt{3}}{4}}dx$
$V = 62.35$

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)

2. Part a, I do not think you're wrong. I get $18\pi$ as well.

Your work and logic seems OK. Same with part b.

3. Estimate with a sphere of radius 3.

$\frac{1}{2}\cdot\frac{4}{3}\pi 3^{3} = 18\pi$

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.

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