Double Integral Setup for a Rectangular Region

[harpazo]

Set up an integral for both orders of integration.
DO NOT EVALUATE THE INTEGRAL.

Let S S = double integrals

Let R = region

S S sinx cos x dA

R: rectangle with vertices (-pi, 0), (pi, 0), (pi, pi/2), (-pi, pi/2)

I am having such a hard time with the set up.

I can graph the region but have no idea how to proceed from there. I need solution steps.

 

[MarkFL]

If we use horizontal strips, then we would use:

$$I=\int_0^{\frac{\pi}{2}}\int_{-\pi}^{\pi} \sin(x)\cos(x)\,dx\,dy$$

If we use vertical strips, then:

$$I=\int_{-\pi}^{\pi} \sin(x)\cos(x)\int_0^{\frac{\pi}{2}} \,dy\,dx$$

 

[harpazo]

If we use horizontal strips, then we would use:

$$I=\int_0^{\frac{\pi}{2}}\int_{-\pi}^{\pi} \sin(x)\cos(x)\,dx\,dy$$

If we use vertical strips, then:

$$I=\int_{-\pi}^{\pi} \sin(x)\cos(x)\int_0^{\frac{\pi}{2}} \,dy\,dx$$

Wonderful but I do not know what limits of integration apply to dxdy as oppossed to dydx.

 

[MarkFL]

Wonderful but I do not know what limits of integration apply to dxdy as oppossed to dydx.

I've given the limits in the definite double integrals. :D

 

[harpazo]

I've given the limits in the definite double integrals. :D

How do you decide which limits to apply for each double integral?

 

[MarkFL]

How do you decide which limits to apply for each double integral?

From the region $D$...in this case it is a rectangular region bounded by the lines $$x=-\pi,\,x=\pi,\,y=0,\,y=\frac{\pi}{2}$$.

 

[harpazo]

From the region $D$...in this case it is a rectangular region bounded by the lines $$x=-\pi,\,x=\pi,\,y=0,\,y=\frac{\pi}{2}$$.

I will practice more similar questions.