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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) $f(x, y)=2 x^{2} y$

(a) $32 / 3$(b) $\frac{2048}{21}$

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 evaluate the double integral of F of X. Y. For the region are which I'll note is bounded below by y equals two X squared Bounded above by the line y equals eight. So for part a we are given that f just equals one. So we'll find that now if we look at this we can see that it's about equally um simple between vertical and horizontal, But I've chosen to treat this as a horizontally simple region. So we're integrating X between zero and 2 and inside of that. We're integrating why? From two X squared up to eight And then we have one dy DX. So evaluating that inner integral, we'd get just um just eight minus two X squared which we're integrating over X. So this becomes eight x minus two X cubed over three, Evaluated from 0 to 2. Which will give us the result. 32/3. Then for part B, I'll use the same order of integration integrating over Why first then over X. So we have 0 to 22 X squared 28 of Now we have f of x Y equals two X squared y dy dx. So, first integrating over why we'll get the results two times X squared times Y squared over two or just X squared y squared, Evaluated from two x squared up to eight which is being integrated over X evaluating that inner term, we would have that the integral and is going to become 64 X squared minus four X to the power of six dx. So this will then become 64/3 X to the power of three minus four X. The power of 7/7, Evaluated from 0 to 2, Which will give us the result 2048 over 21.

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