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Determine the region $R$ determined by the given double integral.$$\int_{0}^{1} \int_{\sqrt{\frac{y}{3}}}^{\sqrt{4-y}} f(x, y) d x d y$$

Bounded by $y=3 x^{2}, y=4-x^{2}$ and $y=0,$ see Fig. Ex. 17

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 6

Double Integrals

Partial Derivatives

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Lectures

12:15

In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

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Determine the region $R$ d…

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Evaluate the double integr…

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for this problem, we are asked to determine the region are determined by the given double integral. So if we look, we can see that our innermost integral, we are integrating over X and we have that. It's going to be between the square root of why over three Up to the Square Root of four -Y. And then from the exterior integral we have that access between zero and 1. Or excuse me? Why is between zero and one? So if we try to plot this out well actually have to one second here, we have X defined as a function of why. But what we can do is figure out that the lower bound would correspond to Y equals three X squared. In the upper bound on X corresponds to let's see here, X squared equals four minus Y. So it would be Y equals four minus x squared. If we plot this out now and see that the region that we're integrating over bounded below by Y equals three X squared bounded above by y equals four X squared to the left by X equals zero and to the right by X equals one.

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