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Numerade Educator



Problem 22 Easy Difficulty

Find each double integral over the rectangular region $R$ with the given boundaries.
\iint_{R}\left(x^{2}+4 y^{3}\right) d y d x ; \quad 1 \leq x \leq 2,0 \leq y \leq 3




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Video Transcript

All right. So we wanted to evaluate. Um, basically, given this function, Z equals X squared. Plus four y to third. Um, we want to evaluate a double air goal over the region. Are, which is given by, um, basically x and y when x and y are when access between one and two. And why is between zero and three? Okay, so basically, this is like a little set. Notation here were saying that our region are is the set of all points X and y such that access between these two numbers on why is between these two numbers? Okay, so let's draw this region are out a little bit. Let's begin idea of what we're doing. You These lines are not street mi mama sick. Okay, You said you this. Yeah. Okay, good. So we have x between one and two. And then why between one and zero and three. So we have this region right here. Okay, so this would be our, um So basically, since we have this function, we're gonna set up the double integral for this function, X squared, plus four wide to the third. And then I think the order of integration doesn't really matter here, so I'm just going to d y the x Um, basically, we're integrating over this region. So how I like to right the limits and how I remember everything is basically since we have what it's d y dx, we always evaluate from the inside out. So we evaluate this integral and then this interval. So when we're looking at or integrated spent two d why we're thinking about how does why range in this case So and this problem wide ranges from 0 to 3 and then looking at DXC outer Enbrel, we know that X ranges from 1 to 2. So where limits are from 1 to 2 and this would be our double integral that we need to evaluate. So no going now going on to evaluating it. We have been a girl from sort of three x squared spore wide, third de y. This is equal to, um, remember actually think about all other variables as Constance. Besides, the variable you're in greater respect to so X in this case would be a constant, so just easy to slap a lie on their and then this should be four wives. The fourth over for using divers. Power rule three. This is equal to X square. Why, plus y to the fourth sort of three plugging in three first, um, remember, nor to know which variable you're playing into. Just think about what bearable you're integrating. Expect to. So in this case would be why? So we're just plugging it back into Why? So we have X squared times three plus three to the fourth is 81 minus. Uh, this whole thing should just be served because both of them have wise here. So three times any number is zero. So this is basically the result of the inside interval. And now playing in playing that into our outside general, get the 1 to 2 a three x squared plus 81 dx. And then this should be, um, reverse power rule again, we have excellent third, because if we derived this, we get this. Say we just bring this three down and then happen to here, which would be the same as both. And then we have plus 81 x from 1 to 2 playing into first, we have eight plus 1 62 and then minus one plus 81. Thank you. And then we get 1 70 minus 82. This is equal to 88 that is our final answer.

Rutgers, The State University of New Jersey
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