# Thread: Question regarding the equation of an ellipsoid

1. ## Question regarding the equation of an ellipsoid

Ellipsoid: $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$

Quick question. I am to take a horizontal cross-section of this shape and use it to find the volume of the entire shape by the slicing method. Area of an ellipse is given, ${\pi}ab$. These are the steps that my prof wrote for me:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1-\frac{z^2}{c^2}$

$\frac{x^2}{a^2(1-\frac{z^2}{c^2})}+\frac{y^2}{b^2(1-\frac{z^2}{c^2})} = 1$

$A(z) = {\pi}(a\sqrt{1-\frac{z^2}{c^2}})(b\sqrt{1-\frac{z^2}{c^2}})$

So $V = {\pi}ab\int_{z=-c}^{z=c}(1-\frac{z^2}{c^2})dz$

When integrated I did obtain the answer of $\frac{4{\pi}abc}{3}$ but I am just not sure how to correctly determine the limits of integration (z=-c to z=c). Thanks!

2. Originally Posted by Em Yeu Anh
Ellipsoid: $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$

Quick question. I am to take a horizontal cross-section of this shape and use it to find the volume of the entire shape by the slicing method. Area of an ellipse is given, ${\pi}ab$. These are the steps that my prof wrote for me:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1-\frac{z^2}{c^2}$

$\frac{x^2}{a^2(1-\frac{z^2}{c^2})}+\frac{y^2}{b^2(1-\frac{z^2}{c^2})} = 1$

$A(z) = {\pi}(a\sqrt{1-\frac{z^2}{c^2}})(b\sqrt{1-\frac{z^2}{c^2}})$

So $V = {\pi}ab\int_{z=-c}^{z=c}(1-\frac{z^2}{c^2})dz$

When integrated I did obtain the answer of $\frac{4{\pi}abc}{3}$ but I am just not sure how to correctly determine the limits of integration (z=-c to z=c). Thanks!
You want to integrate over the entire ellipsoid. The ellipsoid itself extends from z= -c up to z= c.