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Numerade Educator



Problem 31 Hard Difficulty

Find the volume under the given surface $z=f(x, y)$ and above the rectangle with the given boundaries.
z=x^{2} ; \quad 0 \leq x \leq 2,0 \leq y \leq 5




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Video Transcript

Okay, so we're given the surface z equals X squared. And basically, we want to find the volume between this surface on the Z equals zero plane. So what I mean by that is we draw out the X Y c cordant axes. We're trying to find this volume between the Z equals zero plane and this surface. Okay. And we're looking for that of the volume between this surface and this surface for the region described by these sets of points. So it's the seven points X y such that, um, X is between two on wise between zero and five. So this little set notation and basically kind of reads like English, if you know what What is going on? Um basic. Oh, wait. Um, you rewrite this little bit. Think I messed stuff between? Sorry. Give me one moment. Okay. Let me rewrite this. Actually got this s so bad. Okay, One sec. So, basically, our region are is described by the set of all points, which are element of r squared. Such that zero x is between zero and two on. Dwight is between zero and five. So reading this in english, it's basically are, is the set of all points X and Y that air in R squared R squared is basically like the set of all set of all co ordinate pairs such that X is between zero and two and wise between certain five. So that's what our region, ours and our region are kind of looks like a tangles. So you have hi be this rectangular region here that we want to find the volume between this surface on the surface. So in order to do that, we set up a double integral, um, over function X squared. Ta on. And I'm gonna do with u x t y Integral because I just want to get the X out of the way first and now, nor to write the limits correctly. I just think about how each variable is ranging. So since we're enduring respect to experts on the inside, I think about how does X range? Because the region is stated in this way. We have X rating from 0 to 2 and that why ranging from 0 to 5. So this would be our double in April here and just so valuing this, it's very easy to of X squared T X. This is equal to X 1/3 over three. Evaluate from 0 to 2. She ate thirds my zero. And then we want to put this into the outside integral. So 05 of 1/3 t Why and then this should be. There's cereal five t y. And a little trick here is first not very useful here, but I like to do it. But basically, if you're taking the integral of the function, one of any variable doesn't matter if it's d Y d x DZ de road. Do you think that if I whatever variable you want shoes? Basically, this is always gonna be the length of this. So what I mean by that is this new rule is just basically the length of our interval. So it should just be 1/3 times five and usually noodles give you area. But in this case, it gives you the length of the interval and you'll learn later that the double integral of the function one D a over a region are is basically just the area of region are so whenever we're in agreeance a function one it the dimension of our answer always scales down by one. So if we're finding the area right and we're using a single integral, usually you would find the area. But if we're integrating the function one, it basically scales down one dimension. And now we're finding the length of the line and then usually for a double, integral, you're finding the volume. But if you're integrating the function one, you should just give you one dimension down or basically the area of our integration. So 1/3 times five is 40/3 on that is our final answer.

Rutgers, The State University of New Jersey
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