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

(a) 9(b) 9

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 6

Double Integrals

Partial Derivatives

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for this problem we are asked to find the double integral of F of xy over the region R Which is bounded by the lines. X equals zero, Y equals zero and y equals the square root of nine minus x squared mm hmm. And in part our f of X is just X. So what we'll want to do here is first integrate over why and we'll see why in a second. So we'll have integral from 0 to 3 over X. And then from zero to route nine minus X squared over why. And we are integrating X. D X D Y B X. So the result of that inner integral is going to be X times the square root of nine minus x squared dx. Now, in this form, this is an integral that we can evaluate pretty directly specifically, we can have you equals nine minus X squared which is a pretty straightforward integral to do. So let's see here we have D U equals negative two X. Dx. So D X equals negative. D'you over two X. So this will become negative 1/2 times the integral from U. Of zero up to you of three of the square root of U. Or U to the power of one half ew. Which will then become negative 1/2 times 2/3 times U to the power of 3/2, evaluated from U of zero up to you of three. So we can then substitute back in our definition for you. So it's negative one half. Um 2/3. Actually, I'll note that the twos can divide out. So it's just negative 1/3 times nine minus X squared to the power of 3/2 evaluated from 0 to 3 which will give us a final result then of nine. Then for part B. Let's see here for part B we'll want to actually integrate over why first if we do that we'll get that same form that we can easily evaluate with U substitution. So we're still integrating from 0 to 3 then we're integrating or excuse me? We integrated over Whyfors there for part B. We want to integrate over X first. So integrating over X would be integrating from zero up to not route nine minus Y squared rather than nine minus X squared. Then we have dx dy so again this will become integral from 0 to 3 of y times root nine minus Y squared D Y. And this is identical to what we just solved Except for now we're saying why instead of X but that doesn't change what the result will be. So we can see that the integral will be nine again

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