Ok, got another one for you, thanks for all the help, you guys are great!
If the base of a solid is given by the ellipse: (x^2)/4 + (y^2)/9 = 1
then what is it's volume if the solid (cross sectional view) is an isosceles triangle of heigt 2?
Ok, got another one for you, thanks for all the help, you guys are great!
If the base of a solid is given by the ellipse: (x^2)/4 + (y^2)/9 = 1
then what is it's volume if the solid (cross sectional view) is an isosceles triangle of heigt 2?
Refer back to my previous post here.
Just note that the area of the slices will change:
$\displaystyle Area_{\triangle}=\tfrac{1}{2}(base)(height)=\tfrac {1}{2}\left[\tfrac{4}{3}\sqrt{9-x^2}\right](2)=\tfrac{4}{3}\sqrt{9-x^2}$
Thus, $\displaystyle V(x)=\tfrac{8}{3}\int_0^3 \sqrt{9-x^2}\,dx$
Take note of this: To evaluate the integral...make a trig substutition:
let $\displaystyle x=3\sin\vartheta$.
Can you take it from here?
--Chris
EDIT : Woops...I just noticed that the equation changed slightly...
the distance from one side of the ellipse to the other should now be $\displaystyle 3\sqrt{4-x^2}$ (Verify)
Thus, the area of each slice would be $\displaystyle 3\sqrt{4-x^2}$ and the volume would be $\displaystyle 6\int_0^2 \sqrt{4-x^2}\,dx$
You still would need to apply a trig substitution:
let $\displaystyle x=2\sin\vartheta$