# Thread: volume of shape by using perpendicular cross sections?

1. ## volume of shape by using perpendicular cross sections?

the base of a loudspeaker is determined by the two curves y= (x^2 / 10) and y= (-x^2 / 10) for 1< or= x< or= 4. for this loud speaker , the cross sections are perpendicular to the x-axis are squares. what is the volume of the loudspeaker in cubic units? i tried many different ways and keep getting 4.092 .. but the answer is supposed to be 8.184. which is what i get if i multiply it by two.. but why?? HELP.

2. Originally Posted by calcgeek_17
the base of a loudspeaker is determined by the two curves y= (x^2 / 10) and y= (-x^2 / 10) for 1< or= x< or= 4. for this loud speaker , the cross sections are perpendicular to the x-axis are squares. what is the volume of the loudspeaker in cubic units? i tried many different ways and keep getting 4.092 .. but the answer is supposed to be 8.184. which is what i get if i multiply it by two.. but why?? HELP.
We will do it using a little bit of hand-waving but it makes it easier. Break up this solid into many many pieces (cross-sections). Pick a cross section at a point $\displaystyle x$. The width of this cross section is $\displaystyle dx$. While the base length is $\displaystyle \tfrac{x^2}{10} + \tfrac{x^2}{10} = \tfrac{x^2}{5}$. We are told that the cross sections are squares which means that the volume of each cross section is $\displaystyle \left(\tfrac{x^2}{5}\right)^2dx$. Now integrate that from $\displaystyle 1\text{ to }4$.

3. Originally Posted by calcgeek_17
the base of a loudspeaker is determined by the two curves y= (x^2 / 10) and y= (-x^2 / 10) for 1< or= x< or= 4. for this loud speaker , the cross sections are perpendicular to the x-axis are squares. what is the volume of the loudspeaker in cubic units? i tried many different ways and keep getting 4.092 .. but the answer is supposed to be 8.184. which is what i get if i multiply it by two.. but why?? HELP.
side of a square cross-section is $\displaystyle s = \frac{x^2}{10} - \left(-\frac{x^2}{10}\right) = \frac{x^2}{5}$

$\displaystyle V = \int_1^4 \left(\frac{x^2}{5}\right)^2 \, dx = 8.814$

4. dude. thats awesome! thank you! been working on this problem for weeks.. and i keep skipping it! my whole class has been struggling. not even my calc teacher knew. we were starting to say they had made a mistake on the practice AP exam.. =] thx!
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