🤔 Find out what you don't know with free Quizzes 🤔Start Quiz Now! # Find the area of one side of the "wall" standing perpendicularly on the curve $2 x+3 y=6,0 \leq x \leq 6,$ and beneath the curve on the surface $f(x, y)=4+3 x+2 y$.

Integrals

Vectors

Vector Functions

### Discussion

You must be signed in to discuss.
##### Top Calculus 3 Educators ##### Catherine R.

Missouri State University   Lectures

Join Bootcamp

### Video Transcript

Okay, folks. So in this video, we're gonna take a look at problem number 32 um, where were asked to find the area of one side of the wall, um, standing perpendicular early on a curve and beneath the curve, Um, and were given both of those cars. Okay, so, um, this might look like a word problem, but it's really not. It's really just regular line and girl problem. So in order for you to find an area, um, we're going to do basically the same thing that we did for, you know? Uh huh. Like simple two dimensional area finding problems where were given a curve aftereffects and were asked to find it. This this area, you know. So the way you do this kind of problem is I'm pretty sure you already know. You just integrate. Have effects along the thea the infinite decimal ice like Mindy X. Okay, so that's how you do. You know, a basic two dimensional area problem. And now that we're in three d, you know that the idea is really still the same? Um, So we're gonna we're gonna do an integral of F. But now it's not. It's more than just a function effects. It's also a function of why? Because, you know, we're in three D multiplied by instead of DX. We have DS. Okay, so So it's this is really kind of an analogy. Um, And I hope you understand that that even though we're in three D, the idea is still the same, you know, to find the area means to do in general a line, integral. All right. Anyway, that was the analogy. So now let's crank out some eligible. Here we have. Ah, an integral um Well, the function f is going to be four plus three x plus two. Why? So that's the function f multiplied by the Yes, but I'm not gonna write DS. I'm gonna write DS in another way because DS is simply a one plus y prime squared, multiplied by D X. Now, um, let me plug in the integration limits here. No, um, this is gonna be Let me write it this way. We have four plus three X plus two y. What is why prime? Well, why? Prime is first of all, um or that the curve that we're integrating over is two x plus three y equal six. And that means three y equals six minus two x. And why is two minus two you over three X and why Prime is negative to over three. So why Prime Squared is for overnight. Okay, multiplied by D x. No, this thing is someplace collaborative. 13 overnight multiplied by thin to grow of a plus, um, 5/3 X D. X from 0 to 6. Okay. And when you plug in the when you evaluate this thing, you get through 13 overnight, eight x plus 5/6 x squared between zero in six new. This is gonna be equal to route 13 overnight, 48 plus 30. So now we end up with 26 Route 13. And that's it for this view. This is the answer for problem number 32. Thank you for watching. Um, and this This is the answer. Thank you for watching by University of California, Berkeley

#### Topics

Integrals

Vectors

Vector Functions

##### Top Calculus 3 Educators ##### Catherine R.

Missouri State University   Lectures

Join Bootcamp