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Evaluate $\int_{R} \int f(x, y) d A$ for $R$ and $f$ as given.(a) $f(x, y)=1,$(b) $f(x, y)=4 x^{2} y$

(a) $8 / 3$(b) $\frac{1504}{105}$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 6

Double Integrals

Partial Derivatives

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calculate the integral, as…

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Evaluate the integral.

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for this problem we are asked to evaluate the double integral of F of X for r and F. As given where I'll note that the region R is the interior of this sort of piecewise defined area where we have y equals three X squared on the left hand side here and y equals four minus x squared on the right hand side here. Which I'll write alternatively as being x equals the square root of Y over three And x equals the square root of four -Y. Because for part A we're integrating just one over this region. We'll want to treat this as a vertically simple region so we can see that we'd be integrating from three X squared up to four minus X squared dy uh Excuse me. Other way around. We want to use the Definitions that we have down below so it's square root of why over three in as the lower bound and then square root of four minus y is the upper bound and we're integrating over x first. Then our integral for why is just between zero and 3. So we can see that well integrating that first or a evaluating that first integral we'd get just root for minus y minus the square root of Why? Over three. Why? Now that first term is rather simple. I'll just say that it's use substitution U equals four minus y. And then route. Why over root three relatively simple. We can write that as Y to the power of one. Half over three. So we'll find that the result is going to be negative. 2/3. Excuse me? One moment Integral from 0 to 3 of negative. 2/3 times four minus y. The power of 3/2 minus 2/3 times. Wides the power of 3/2 over root three. Dy now or excuse me, should have been D y d X. Excuse me. One moment here. Made a mistake somewhere. Uh My silly mistake here that actually that we've already evaluated both integral. This is just being evaluated from 0 to 3 Which will give us a result. 8/3 then for part B will be integrating over the same region, which I'll define in the same order of integration. But now our into grand is for X squared why? Dx Dy So this is going to end up being a little bit ugly no matter how we do it. But we'll have that evaluating that inner integral, we would get 4/3 times X cubed times y evaluated from root y minus three Up to or not? Why -3. Excuse me? It's why over three Up to Route four -Y integrated over why? So our integrated now will become let's see here. So it will be 16/3 times root four minus y minus 4/3 times y squared times the square root of four minus y -4/9 root three Times Why? to the power of 5/2. Dy Now integrating that. So this is going to be a little bit brutal to write out, but we'll get negative. 1,024 over 105 times the square root of four minus y -128 over 105 times. Why root for minus Y Plus 2 56 Over 1. 5 times y squared root for -Y. That's four minus y, not four minus y squared -8/21 times the square root of four minus 1. 8/21. Y cubed root for minus why minus eight Y to the power of 7/2, Divided by 63 root three, Evaluated from 0 to 3, which will give us our final result here. 1504 over 105

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