- #1

mr_coffee

- 1,629

- 1

Hello everyone, I'm having issues figuring out how you find the bounds of each of these graphs, I have the solution manual but I still don't see how they did it. For example:

The directions say to evaluate the double integral:

http://img241.imageshack.us/img241/903/scan0001nx8.jpg [Broken]

On the first image, for #15.

THe directions say: doulbe integral D y^3 dA. D is the triangluar region with vertices (0,2) (1,1) (3,2)

from (0,2) to (1,1) they got the equation of the line to be:

x = 2-y;

and from

(1,1) to (3,2) they got

x = 2y-1;

So I used the line formula, y = mx + b;

b is the intercept of the y-axis:

m = (y1-y2)/(x1-x2);

m = (1-2)/(1-3) = -1/-2 = 1/2;

y = 1/2x + 2

y-2 = 1/2x

x = 2y-4

They also got bounds 1 to 2 for the dy, and I'm also not sure how they came up with that either.

But this isn't the answer, and also for 17, 19 and 21 i don't see how they are getting these bounds...any help would be great!

For #17, i see how they get y = sqrt(4-x^2) and y = -sqrt(4-x^2) is it dy bounds becuase its in the y -axis and the dx bounds is -2 to 2 becuase in the x direction the radius is 2?

but for #19 i don't see how they are getting x^4, or x, i see from (0,0) to (1,1) that's the distance of 1, so I'm guessing that's y the dx is 0 to 1?

The directions say to evaluate the double integral:

http://img241.imageshack.us/img241/903/scan0001nx8.jpg [Broken]

On the first image, for #15.

THe directions say: doulbe integral D y^3 dA. D is the triangluar region with vertices (0,2) (1,1) (3,2)

from (0,2) to (1,1) they got the equation of the line to be:

x = 2-y;

and from

(1,1) to (3,2) they got

x = 2y-1;

So I used the line formula, y = mx + b;

b is the intercept of the y-axis:

m = (y1-y2)/(x1-x2);

m = (1-2)/(1-3) = -1/-2 = 1/2;

y = 1/2x + 2

y-2 = 1/2x

x = 2y-4

They also got bounds 1 to 2 for the dy, and I'm also not sure how they came up with that either.

But this isn't the answer, and also for 17, 19 and 21 i don't see how they are getting these bounds...any help would be great!

For #17, i see how they get y = sqrt(4-x^2) and y = -sqrt(4-x^2) is it dy bounds becuase its in the y -axis and the dx bounds is -2 to 2 becuase in the x direction the radius is 2?

but for #19 i don't see how they are getting x^4, or x, i see from (0,0) to (1,1) that's the distance of 1, so I'm guessing that's y the dx is 0 to 1?

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