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Sketch the region of integration for the following integral. ∫π/40∫6/cos(θ)0f(r,θ)rdrdθ
1 answer:
Answer:
The graph is sketched by considering the integral. The graph is the region bounded by the origin, the line x = 6, the line y = x/6 and the x-axis.
Step-by-step explanation:
We sketch the integral ∫π/40∫6/cos(θ)0f(r,θ)rdrdθ. We consider the inner integral which ranges from r = 0 to r = 6/cosθ. r = 0 is located at the origin and r = 6/cosθ is located on the line x = 6 (since x = rcosθ here x= 6)extends radially outward from the origin. The outer integral ranges from θ = 0 to θ = π/4. This is a line from the origin that intersects the line x = 6 ( r = 6/cosθ) at y = 1 when θ = π/2 . The graph is the region bounded by the origin, the line x = 6, the line y = x/6 and the x-axis.
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Remark
This will be a little late, but I'll answer it anyway. The more you see of the sine law, the clearer it will be.
Step One
Set up the sine Law.
Step Two
Substitute values
Sin(x) = ??
sin(75) = 0.9659
Step Three
Cross multiply and Solve
22* Sin(x) = 12 * 0.9659
22* Sin(x) = 11.591 divide by 22
sin(x) = 11.591 / 22
sin(x) = 0.5269
Step Four
You have to know how to use the inverse of an angle to get the answer. I do it as my calculator would do it. If it doesn't work with yours, let me know.
x = sin^-1(0.5269)
2nd F
Sin^-1
(
0.5269
)
=
You should get 31.79 degrees.