How do you find the volume of the solid with base region bounded by the curve #9x^2+4y^2=36# if cross sections perpendicular to the #x#-axis are isosceles right triangles with hypotenuse on the base?

1 Answer
Wataru
Sep 12, 2014

The volume of the solid can be found by
#V=1/4int_{-2}^2(36-9x^2)dx=24#.

Let us look at some details.
If we rewrite #9x^2+4y^2=36# as (by dividing by 36)
#x^2/2^2+y^2/3^2=1#,
we realize that it is en ellipse that spans from #x=-2# to #x=2#.

By solving for #y#, we have
#y=pm1/2sqrt{36-9x^2}#
Since the area #A(x)# of the cross-sections can be expressed as

#A(x)=1/2#(base)(height)
#=1/2 cdot sqrt{36-9x^2}cdot 1/2sqrt{36-9x^2}#
#=1/4(36-9x^2)#

So, the volume of the solid can be found by
#V=1/4int_{-2}^2(36-9x^2)dx#
by symmetry about the y-axis,
#=1/2int_0^2(36-9x^2)dx#
#=1/2[36x-3x^3]_0^2=1/2(72-24)=24#

Related questions
See all questions in Calculating Volume using Integrals
Impact of this question
9804 views around the world
You can reuse this answer
Creative Commons License