Suppose we want to integrate

Question

Sometimes we switch coordinates because it makes the integrand easier. Let D� denote the triangular region with vertices at (0,0),(1,0) and (0,1). This region is not round. However, suppose we want to integrate

integral integral D 4(x+y)^2 /x^2+y^2 dA

cylindrical coordinates for this are 4+8cos(t)sin(t) t = theta

Next: we need to describe this triangle using polar coordinates. Sketch the triangle. What is the equation of the line bounding this region (with the axes) in the first quadrant? y=

Can you walk through how to find what y equals. I'm having a hard time visualizing it and trying to come up with a formula.

Answer 1

Answer 2

Here's one way to find r(θ) equation for y=-x+1

rsin(θ)=-rcos(θ)+1

r=1/(cos(θ)+sin(θ))

r=(cos(θ)-sin(θ))/cos(2θ) ,,, 0<θ<π/2

note, when θ=π/4, r=(√2)/2

The integral evaluated r=0 to r=1 and θ=0 to π/2

for volume under curve over the domain =π+2

Answer 3

It's the straight line passing through points (1, 0) and (0, 1).

slope = rise / run = -1/1 = -1

y-intercept = 1, this is the point (0, 1)

y = (slope)x + (y-intercept) = - x + 1 = 1 - x

Answer 4

"""

Given the three vertices, (0, 0), (1, 0), and (0, 1),

we can connect them with equations.

Since (0, 0) and (0, 1) are both on the y-axis, the

equation that connects them is simply x = 0. Similarly,

since (0, 0) and (1, 0) are both on the x-axis, the

equation that connects them is simply y = 0.

[Note, these equations are true for all values of y

and x, respectively, but since we're only considering

them between two point they should really be bounded

by the points, 0<= y <= 1 and 0 <= x <=0, respectively,

however we don't emphasize that here.]

To find the equation that connects the point (1, 0) and

(0, 1) we must involve a bit more algebra. Namely, the

point-slope intercept form of a linear equation:

y - y1 = m * (x - x1), where m = (y2 -y1) / (x2 - x1)

and (x1, y1) and (x2, y2) are some points on the line.

Let (x1, y1) = (0, 1), that is, x1 = 0 and y1 = 1, and

(x2, y2) = (1, 0), that is, x2 = 1 and y2 = 0. Then,

substituting these values into the equation:

y - y1 = m * (x - x1),

y - y1 = [ (y2 -y1) / (x2 - x1) ] * (x - x1),

y - (1) = [ (0 - 1) / (1 - 0) ] * (x - 0),

y - 1 = [ -1 / 1 ] * x,

y - 1 = -1 * x,

y = -x + 1.

I have put all of this together in Desmos for you as well:

https://www.desmos.com/calculator/hihrsrx62i

Please feel free to play with the Desmos file and modify

it as necessary. Let me know if you need further help with

the question.

All the best!

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