Kind of ambiguous question not entirely sure what she's asking (Calc 2).)

[arishorts]

Homework Statement

"The base of a solid is the region enclosed by y=4-x^2 and the x-axis. Find the volume of the solid if cross-sections perpendicular to the y-axis are isosceles right triangles with hypotenuse in the base"

I think she forgot to add to rotate the equation around the y-axis, but i'll let you be the judge of that.

Homework Equations

y=4-x^2
V= a∫b A(x)dx

The Attempt at a Solution

I've been staring at my textbook for the past hour and a half and I have no clue where to begin.EDIT:
think i figured out the answer. please correct me if I'm wrong.

I started by constructing the cross-sections by splitting an isosceles triangle down the middle, giving me two more isosceles triangles. More importantly, it allowed me to calculated the formulas for the height and base of the cross section:

A(x)= base*height*1/2
A(x)=(2*rad(4-x^2))*(rad(4-x^2))*(1/2)
=(4-x^2)

-2∫2 (4-x^2)dx= 2* 0∫2 (4-x^2)dx=
2(4x+(x^3/3) integrand [i believe this is the right term?] from 0 to 2)
2(8+(8/3))=
16 +(16/3)=64/3!

survey says?

 

[arishorts]

think i figured out the answer. please correct me if I'm wrong.

I started by constructing the cross-sections by splitting an isosceles triangle down the middle, giving me two more isosceles triangles. More importantly, it allowed me to calculated the formulas for the height and base of the cross section:

A(x)= base*height*1/2
A(x)=(2*rad(4-x^2))*(rad(4-x^2))*(1/2)
=(4-x^2)

-2∫2 (4-x^2)dx= 2* 0∫2 (4-x^2)dx=
2(4x+(x^3/3) integrand [i believe this is the right term?] from 0 to 2)
2(8+(8/3))=
16 +(16/3)=64/3!

survey says?

 

[LCKurtz]

Your cross sections are given perpendicular to the y axis. It looks like you have set it up as though the cross sections are perpendicular to the x axis. You need a dy integral.