General Slicing Method

[Starblazer]

The problem

Use the general slicing method to find the volume of the following solid.

The solid with a semicircular base of radius 7 whose cross sections perpendicular to the base and parallel to the diameter are squares.

My solution so far

$\displaystyle \int\limits_a^b {A(x)dx}$

area of a square = Length x Width

semi-circular base $\displaystyle y = \sqrt {49 - {x^2}}$

To get A(x) of squares
$\displaystyle A(x) = {(\sqrt {49 - {x^2}} )^2} = 49 - {x^2}$

$\displaystyle \int\limits_{ - 7}^7 {(49 - {x^2}} )dx$

$\displaystyle = 49x - \frac{{{x^3}}}{3}|_{ - 7}^7$

$\displaystyle = \left[ {49(7) - \frac{{{{(7)}^3}}}{3}} \right] - \left[ {49( - 7) - \frac{{{{( - 7)}^3}}}{3}} \right]$

$\displaystyle = \frac{{1372}}{3}$ units^3

The correct answer is $\displaystyle = \frac{{2744}}{3}$ which is double my answer
Can you tell me what is wrong with my solution?

 

[HallsofIvy]

There is nothing wrong with your answer. The "correct answer" you give is for a full circular base.

 

[flyingsquirr3l80]

Late response, but I hope it helps someone encountering the same problem like myself.

Consider for a moment a full circle with its center at the origin. The height of the circle is sqrt(49-x^2). In other words, substituting 0 gives you a height of +7 or -7. Therefore, the full height of the circle is 2*sqrt(49-x^2). Since the full height of the circle represents a side of the square, we can derive the area: A = (2*sqrt(49-x^2)(2*sqrt(49-x^2) or A = 4(49-x^2). Because we interested in a semicircle in this case, the limits of integration are (0,7). The final answer is indeed 2744/3 cubic units.