# Thread: volumes of solids using integrals

1. ## volumes of solids using integrals

if the base of a solid is a square perpendicular to the x-axis enclosed by the ellipse
(x^2)/9 + (y^2)/4 = 1

Find it's volume

Thanks!

2. Originally Posted by winterwyrm
if the base of a solid is a square perpendicular to the x-axis enclosed by the ellipse
(x^2)/9 + (y^2)/4 = 1

Find it's volume

Thanks!
We are finding the volume using the technique of slices.

Recall that $\displaystyle V(x)=\int_a^b A(x)\,dx$, where $\displaystyle A(x)$ is the area of each slice.

To find out the area of each slice, we need to know the length of one side.

Since the region is an ellipse:

we can figure out that the distance between each side of the ellipse could be defined as $\displaystyle S(x)=4\sqrt{1-\frac{x^2}{9}}=\tfrac{4}{3}\sqrt{9-x^2}$

Thus, the area of each square slice will be $\displaystyle A(x)=(S(x))^2=\tfrac{16}{9}(9-x^2)$

Thus, the volume of the solid would be $\displaystyle V(x)=\tfrac{16}{9}\int_{-3}^3 (9-x^2)\,dx=\tfrac{32}{9}\int_0^3 (9-x^2)\,dx$.

Does this make sense? If you have a question, feel free to ask.

--Chris

3. Thanks, took me a while to catch up to everything you did, but I definitely get it now, I just messed up in the multiplication, instead of simplifying like you did, I mixed up numbers, (I do that alot) and I have the answer, but only the number, so you can understand my frustration, I suppose.