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Problem 55

Use the region $R$ with the indicated boundaries …

Problem 54

Use the region $R$ with the indicated boundaries to evaluate each double integral.
$$\iint_{R} x^{2} y^{2} d x d y ; \quad R$ bounded by $y=x, y=2 x, x=1$$




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Video Transcript

All right. So here we are, integrating disfunction over this specified region functions X squared. Why squared t x t? Why in the region is bounded between, understand? Sketch it quick. So x y No. Okay, so it says y cause X years. Why? Cause X Y equals two accidents be appear next, and then X equals one. So here's our region right here. Okay. And so we're going to make a choice whether we really want to do X first or wife first, and, well, it seems like to me, you should do Why First, if you do x first, you're gonna have to break it up into two equal. So that immigrating X here, it's going to depend on the wires. Ah, Why? From zero one or wanted to. So, I mean, if you if you are not following, just you can just do it both ways and and see what I mean. Okay, so I'm going to integrate, actually. Why? First and now that means I'm going to be fixing an X value between here. Okay. My region is going from zero toe one and X and then where am I going? I'm going from Well, Michael's ex. So in terms of Ana indexing next value. And so I'm going from Why all the way up, Tio? Well, why was to expect in terms of eggs, That's why over too. So this is gonna be exes going from zero to one. And then why is going to be going from X here? Yeah, yeah, Just sex to do X. Okay. And then we have our function X squared y squared. Tio, I k'nex Okay? Yes, over here. And just wise made from eggs up to two ex for any fixed X. So here. No one and I derivative with respect. Why? So we'LL get one third no X squared. Why? Cubes evaluated from Exeter to X. What? What is that there? Okay, so that's I'm gonna cube this and then subtract this cube. So that's looking like So that was just going to be executed within. I ate excused minus one. Excuse us seven execute. But then I have Miss X. Uh, squared. OK, so all in all, it should be seven x to the fifth. Okay, so we have seven thirty compactor out X to the fifth. That's one sixth next to the sixth about a J from zero to one. But this is just one. So we end up with seven over eighteen