The upper limit of integration for r is wrong. The region over which integration is taking place is the triangle bounded by the y axis, the line y = x, and the line y = 6. Along this horizontal line, r = 6 cscΘ.soccerboy10 said:Homework Statement
6...y
∫...∫ x dx dy
0 ...0
The Attempt at a Solution
for reference: π=pi
6...y
∫...∫ x dx dy
0 ...0
y=6=x=6. rcosΘ=6 r=6/cosΘ
goes to
π/2 6secΘ
∫ ...∫... r cosΘ r dr dΘ
π/4...0
soccerboy10 said:π/2.....r=6secΘ
∫ ... [((r^3)/3)cosΘ] dΘ
π/4......r=0
π/2
∫ ...72 sec^2 Θ dΘ
π/4
72tan(π/2) - 72tan(π/4)
last i checked, tan(π/2) was undefined. where did i undergo an error?
The purpose of changing cartesian integrals to polar integrals is to simplify the integration process and make it easier to solve certain types of problems, particularly those that involve circular or symmetric regions.
To convert a cartesian integral to a polar integral, you must first identify the equation of the curve in polar coordinates. Then, you can substitute the appropriate polar variables (r, θ) into the integral and use the appropriate conversion formulae to change the limits of integration.
Polar coordinates are particularly useful for integrating over circular or symmetric regions, as they can simplify the integrand and make it easier to solve. They also allow for a wider range of integration techniques, such as using symmetry or the polar substitution method.
One limitation of using polar coordinates for integration is that they are not as intuitive as cartesian coordinates, and may require more practice to understand and use effectively. Additionally, not all regions or curves can be easily described or integrated using polar coordinates.
No, not all cartesian integrals can be converted to an equivalent polar integral. Some integrals may be more complicated or impossible to express in terms of polar coordinates, while others may not benefit from the conversion and may be easier to solve in cartesian form.