# Find the volume V of the described solid S. The base of S is an elliptical region with boundary curve 9x^2 + 16y^2 = 144. Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base.

1
by jothisaran

2014-10-24T12:20:45+05:30
The area of an Isosceles right triangle is, when given the length of the diagonal, d =
A = 1/4 * d²
The volume of any solid with a cross section of an isosceles right triangle is just the integral of that area along the length of that solid. Because our triangles lie perpendicular to the x-axis, we need to integrate the area along the x-axis to find the volume. So
V = integral of ( 1/4 * d² ) from (x = x1) to (x = x2) dx
The bounds of our integral x1 and x2 we can find by just plugging in y = 0 to our elliptical formula and getting x= - 4 and x = 4. Thus
V = integral of ( 1/4 * d² ) from (x = - 4) to (x = 4) dx
But we still need to express the length of the hypotenuse of our triangle, d, in terms of the parameter that we are integrating, in this case that parameter is x.
Solving this equation for y, we see that y = ( 9 - 9/16 * x2 )
But, if we graph out the ellipse, we can see that the coordinate "y" is only half of the total length of the hypotenouse. So y = d/2, or d = 2y
Now plug in y in terms of xd = 2 * ( 9 - 9/16*x2 )
Now lets plug this in for V.
V = integral of ( 1/4 * (2*( 9 - 9/16*x2 ))² ) from (x = - 4) to (x = 4) dx
or V = integral of ( 9 - 9/16*x2 ) from (x = - 4) to (x = 4) dx
which if you do the arithmatic you get
V = 48