Double integrals in the first octant

[Idealistic]

Find the volume of a solid in the first octant bounded by z = 16 - x[sup:1fkre2xu]2[/sup:1fkre2xu] and the plane y = 5.

set it up like so:

(from 0 to 5)(from 0 to 4)int(16 - x[sup:1fkre2xu]2[/sup:1fkre2xu])dxdy

My question is, how do I integrate in terms of y (after x) when y is not defined in the equation?

couldn't I just do:

5*(from 0 to 4)int(16 - x[sup:1fkre2xu]2[/sup:1fkre2xu])dx ?

 

[galactus]

Your have the correct set up for your integral.

After integrating w.r.t x, we have 128/3.

Now, add a y. The opposite as if you differenatiated. See?.

\(\displaystyle \frac{128}{3}\int_{0}^{5}dy=\frac{128}{3}y\large|_{0}^{5}\)

or reverse the order of integration

\(\displaystyle \int_{0}^{4}\int_{0}^{5}(16-x^{2})dydx\)

\(\displaystyle 5\int_{0}^{4}(16-x^{2})dx\)

YEP!!!. You can do that....Good.

 

[Idealistic]

Oh wow. I slept on that one.

Again, thanks for the clearification.

 

[galactus]

You were correct. That was good. Just two different ways to go about it.