1. Sketch the region of integration & reverse the order of integration. Double integral y dydz... 1st (top=1, bottom =0)... 2nd(inner) integral (top=cos(piex), bottom=(x-2)...
2. Evaluate the integral by reversing the order of integration. double integral sqrt(2+x^3) dxdy... 1st integral (top=1, bottom=0)... 2nd (interior) integral(top=1, bottom=sqrt(y)).
3. Evaluated the integral using a change of variables; double integral sin(x+y) dxdy... below both integral signs is R... over the region R being on the disc x^2 +u^2<=2.
Thanks!
1. To sketch the region of integration, we need to visualize the given limits of integration for each variable. The first integral suggests that the variable of integration for the outer integral is "y" and the inner integral's variable is "x". The limits for the outer integral are from 0 to 1 for "y".
Now, let's focus on the inner integral. The limits for "x" are given as cos(piex) to (x-2). Since it is difficult to visualize the given limits directly, we can use algebraic techniques to simplify them.
First, we set cos(piex) equal to (x-2) to find the intersection points between the two curves. Solving this equation will give us the x-values at which the curves intersect. Next, we substitute these x-values back into either the equation cos(piex) or (x-2) to find the corresponding y-values.
Once we have these intersection points, we can sketch the region of integration on a graph by connecting these points and shading the area inside the region.
To reverse the order of integration, we need to write the double integral with respect to "x" as the outer integral and with respect to "y" as the inner integral.
So, the reversed form of the integral would be double integral (cos(piex) ≤ y ≤ (x-2), 0 ≤ x ≤ 1) y dydx.
2. To evaluate the integral by reversing the order of integration, we start by swapping the order of integration. In the original integral, "x" is the outer integral's variable, and "y" is the inner integral's variable.
The limits for the outer integral are from 0 to 1 for "x", and for the inner integral, the limits are from 1 to sqrt(y) for "y".
Therefore, the reversed form of the integral would be double integral (0 ≤ x ≤ 1, 1 ≤ y ≤ sqrt(y)) sqrt(2+x^3) dxdy.
Now, you can evaluate this integral by integrating with respect to "x" first using the limits given for "x" and then integrating with respect to "y" using the limits given for "y".
3. To evaluate the integral using a change of variables, we need to apply a substitution. Let's choose the substitution:
u = x + y
v = x - y
First, we find the Jacobian determinant of the transformation by taking the partial derivatives of u with respect to x and v with respect to y. The determinant is 2.
Next, we need to find the new limits of integration in terms of u and v. We are given that the region R lies on the disc x^2 + u^2 ≤ 2. We substitute the given equation into our transformation equations:
x^2 + u^2 ≤ 2
(x + y)^2 + u^2 ≤ 2
u^2 + v^2 + 2xy + u^2 ≤ 2
Simplifying, we have:
2x^2 + 2y^2 + 2xy ≤ 0
x^2 + y^2 + xy ≤ 0
x^2 + y^2 + xy ≤ 0
This inequality represents the region R in the u-v plane. By finding the limits of integration for u and v based on this inequality, we can then evaluate the double integral:
double integral (over R) sin(x + y) dxdy
double integral (over the region in the u-v plane) sin(u) * |Jacobian determinant| dudv
double integral (over the region in the u-v plane) sin(u) * 2 dudv
You can evaluate this integral by integrating with respect to u first using the limits derived from the inequality, and then integrating with respect to v using the limits derived from the inequality as well.