Explanation Given: The surface is z = 4 − x 3 − y 3 and rectangular region as 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 Approach: If there is a surface z define as z = f ( x , y ) and region R as a ≤ x ≤ b , c ≤ y ≤ d , then volume ( V ) under given surface and over the rectangular region R is V = ∬ R f ( x , y ) d x d y = ∫ c d ∫ a b f ( x , y ) d x d y Calculation: Given: surface z = 4 − x 3 − y 3 and region 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 Volume under given surface and over rectangular region is

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Chapter 9.5, Problem 4YT

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The volume under the surface z=4−x3−y3 and above the rectangular region R 0≤x≤1,0≤y≤1.

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Chapter 9.5, Problem 3YT

Chapter 9.5, Problem 5YT

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(b) Find the (instantaneous) rate of change of y at x = 5. In the previous part, we found the average rate of change for several intervals of decreasing size starting at x = 5. The instantaneous rate of change of fat x = 5 is the limit of the average rate of change over the interval [x, x + h] as h approaches 0. This is given by the derivative in the following limit. lim h→0 - f(x + h) − f(x) h The first step to find this limit is to compute f(x + h). Recall that this means replacing the input variable x with the expression x + h in the rule defining f. f(x + h) = (x + h)² - 5(x+ h) = 2xh+h2_ x² + 2xh + h² 5✔ - 5 )x - 5h Step 4 - The second step for finding the derivative of fat x is to find the difference f(x + h) − f(x). - f(x + h) f(x) = = (x² x² + 2xh + h² - ])- = 2x + h² - 5h ])x-5h) - (x² - 5x) = ]) (2x + h - 5) Macbook Pro

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The volume under the surface z = 4 − x 3 − y 3 and above the rectangular region R 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 .

Solution Summary: The author explains that the volume under the surface and over the rectangular region is rllV.

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To find:

The volume under the surface z=4−x3−y3 and above the rectangular region R 0≤x≤1,0≤y≤1.

Want to see the full answer?

Chapter 9 Solutions

The volume under the surface z = 4 − x 3 − y 3 and above the rectangular region R 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 .

Solution Summary: The author explains that the volume under the surface and over the rectangular region is rllV.

Videos

To find: The volume under the surface z=4−x3−y3 and above the rectangular region R 0≤x≤1,0≤y≤1.

Want to see the full answer?

Chapter 9 Solutions

To find:

The volume under the surface z=4−x3−y3 and above the rectangular region R 0≤x≤1,0≤y≤1.