Double integral sketching of D = x^2+1 <= y <= 2(x+2)

[justanca]

Hello, I am trying to solve this exercise:

2. Evaluate the following integral over the given region D, sketching this region:

\(\displaystyle \displaystyle\int\int_D\, (x^2\, +\, y)\, dx\, dy,\, \mbox{ where }\, D\, =\, \left\{\, (x,\, y)\, \bigg|\, x^2\, +\, 1\, \leq\, y\, \leq\, 2(x\, +\, 2)\, \right\}\)

I am not really sure how to begin. I think I have to use a cos and sin representation of x and y.

If anyone could give me a hint on how to start, I will try to solve it, because that's the issue at the moment.

Thank you!

 

[Ishuda]

Hello, I am trying to solve this exercise:

2. Evaluate the following integral over the given region D, sketching this region:

\(\displaystyle \displaystyle\int\int_D\, (x^2\, +\, y)\, dx\, dy,\, \mbox{ where }\, D\, =\, \left\{\, (x,\, y)\, \bigg|\, x^2\, +\, 1\, \leq\, y\, \leq\, 2(x\, +\, 2)\, \right\}\)

I am not really sure how to begin. I think I have to use a cos and sin representation of x and y.

If anyone could give me a hint on how to start, I will try to solve it, because that's the issue at the moment.

Thank you!

First make a sketch of
y = x² + 1
This splits the plane into two regions. In which region is
\(\displaystyle x^2\, +\, 1\, \le\, y\).
Another way to say that is, in what region is y unbounded (what region can y go to infinity)?

Now do the same for
\(\displaystyle y\, \le\, 2(x\, +\, 2)\)
except it is y less than or equal to, i.e. y is bounded.

Putting the two together tells you what the region D is and over what you are integrating. In this case, it is a simple region in that the curves intersect in only two places. That is
\(\displaystyle x^2\, +\, 1\, =\, 2(x\, +\, 2)\).
has two solutions and that will give you the bounds to integrate.

 

[justanca]

Thanks, Ive managed to solve it. The result was 90 (although Im not sure abt the final calculations). It's a lot to type but might take a picture of it later. I got 2 points, (3,10 and (-1,2) therefore x<=3 and x>=-1 and x^2+1<=y<=2(x+2).

When integrating (x^2+y)dy, the correct result is x^2*y+y^2/2, right? Treating x as if it's a constant.