I'm having trouble reversing the order of integration of ∫∫dxdy from a=0 to b=2(3)^(1/2) for x and c=y^(2/6) to d=(16-y^2)^(1/2) for y.
I graphed the region of integration and that still doesn't really help me.
i got approximately 7.9 for the ∫∫dxdy which im pretty sure is right.
please help
I'm having trouble interpreting your variables and limits.
c,d are the limits on dy? That's odd, since I'd expect functions of y to be limits on dx. Anyway, interpreting it as
∫[0,√12]∫[y^(1/3),sqrt(16-y^2)] dx dy I get 15.8, twice your answer
I tried to get a handle on the region using the excellent graphing tools at rechneronline dot de slash function-graphs, and it appears the region is to the right of the curve y=x^3 and inside the circle y=sqrt(16-x^2) for x in [0,√12]. Strange limits, since they don't appear to correspond to any useful portion of the picture.
??
To reverse the order of integration for the given double integral ∫∫dxdy, where the limits of integration are a=0 to b=2(3)^(1/2) for x and c=y^(2/6) to d=(16-y^2)^(1/2) for y, we need to visualize the region of integration and come up with new limits of integration.
Here's how you can proceed:
1. Start by graphing the region of integration. Since you mentioned that it didn't help you, let's try another approach.
2. Rearrange the limits of integration to visualize the region better. For y, c=y^(2/6) can be rewritten as y=(6c)^(3/2), and d=(16-y^2)^(1/2) can be rewritten as y=±sqrt(16-d^2). Thus, the limits of integration for y become y=(6c)^(3/2) to y=±sqrt(16-d^2).
3. Next, focus on the limits of integration for x. Since the limits of integration for x are not given explicitly, we will have to find them based on the region of integration.
4. To determine the limits of integration for x, consider the range of x-values that correspond to each y-value within the given limits. Start with the lower limit of y, y=(6c)^(3/2). Notice that for each y-value in this range, the corresponding x-values go from 0 to the x-coordinate of the curve defined by y=y(x). Therefore, the x-limits of integration become x=0 to x=f1(y), where f1(y) represents the equation that defines the curve y=y(x).
5. Now, consider the upper limit of y, y=±sqrt(16-d^2). Similar to the previous step, observe that for each y-value in this range, the corresponding x-values go from 0 to the x-coordinate of the curve defined by y=y(x). Therefore, the x-limits of integration become x=0 to x=f2(y), where f2(y) represents the equation that defines the curve y=y(x).
6. Finally, reverse the order of integration by switching the order of the variables dxdy to dydx. The new limits of integration become ∫(from c to d) ∫(from 0 to f1(y)) dxdy + ∫(from -d to d) ∫(from 0 to f2(y)) dxdy.
7. Now, you can proceed to solve the integral using the new limits of integration and obtain the result. You mentioned that you obtained approximately 7.9 as the value of the integral, so you can verify if this is correct.
Note: The step of finding the equations f1(y) and f2(y), which define the curves y=y(x) for the given limits of integration, might involve further calculations or solving equations specific to the problem.