Starblazer
New member
- Joined
- Mar 28, 2013
- Messages
- 18
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?
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?