6.1.3-eg-1 - s 2 + s 2 = d 2 , so that 2 s 2 = d 2 , or one...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Example Suppose a solid lies between the planes perpendicular to the x -axis at x = - 2 and x = 3. The cross sections perpendicular to the x -axis and these planes run from y = - x + 2 to y = x + 2. Find the volume of the solid if the cross sections are squares with diagonals in the xy -plane. (a) Find the area of a cross section. Since the cross sections are squares whose diagonal d runs from y = - x + 2 to y = x + 2, it follows that d = x + 2 - ( - x + 2) = 2 x + 2 . Recalling the relationship between the side s of a square and its diagonal d , we can use either the Pythagorean theorem
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: s 2 + s 2 = d 2 , so that 2 s 2 = d 2 , or one can use right triangle trigonometry to see that cos 4 = s d , i.e. 2 2 d = s . Thus the area of a cross section is given by A ( x ) = s 2 = ( 2 x + 2) 2 = 2( x + 2) (b) Find the volume of the region. Since the cross sections are stacked along the x-axis from x =-2 to x = 3, the volume V is just the integral of the area A ( x ) from x =-2 to x = 3, i.e. V = Z 3-2 2( x + 2) dx = ( x 2 + 4 x ) 3-2 = 21-(-4) = 25 ....
View Full Document

Ask a homework question - tutors are online