Like

Report

Find the volume under the given surface $z=f(x, y)$ and above the rectangle with the given boundaries.

$$

z=\sqrt{y} ; 0 \leq x \leq 4,0 \leq y \leq 9

$$

72

You must be signed in to discuss.

Johns Hopkins University

Oregon State University

University of Nottingham

Boston College

{'transcript': "Okay, so you're finding the volume under this region. But we can go ahead and set up the double integral. And we're looking at the region in the X y plain X going from zero two four. And why going from zero tonight? And the function we want the volume under is just squared of why me? And you could set up Steve by the ex. But it doesn't really matter which shorter we do it. And so let's do an anti derivative of rea y. So that's just gonna be why to the three hives over three have so dividing by three houses, same thing is multiplied by two thirds. So we just get a factor of two thirds and then we have why to the three has evaluated from zero to nine and then this is just two thirds. Well, he evaluates, uh, nine to the three halves. That's going to be twenty seven, just twenty seven, of course, value mating and zero to zero. And then, well, this is just a constant. So when we integrate underneath this constant twenty seven, we're just going to get a factor of four. Okay, so all in all we get two thirds times twenty seven time's floor. And what is this? OK, so we can cancel one factor of three. So we just have a nine. So it looks like two times. Four times nine, eight times nine. I just need to"}