1. ## ∫∫y^3dA

I hate math...I can't solve even the easiest questions. Setting these things up is a pain, and then integrating is tedious and I almost always make mistakes

ugh...here goes:

∫∫y^3 dA where D is (0,2), (1,1), and (3,2)

I drew a diagram and thought that splitting up the triangle into two parts D1 and D2 and then adding the area of those two would be easiest.

This is what I did:

D1: ∫∫y^3 dxdy where 0 < x < 2-y ; 1 < y < 2 ; I integrate and get 13/10
D2: ∫∫y^3 dxdy where 1 < x < 2y-1 and 1 < y < 2; I integrate and get 49/10.

Adding the two areas I get 31/5 which is perfect because the answer is 147/20.

Now please excuse me while I go kill myself.

2. Originally Posted by crabchef
I hate math...I can't solve even the easiest questions. Setting these things up is a pain, and then integrating is tedious and I almost always make mistakes

ugh...here goes:

∫∫y^3 dA where D is (0,2), (1,1), and (3,2)

I drew a diagram and thought that splitting up the triangle into two parts D1 and D2 and then adding the area of those two would be easiest.

This is what I did:

D1: ∫∫y^3 dxdy where 0 < x < 2-y ; 1 < y < 2 ; I integrate and get 13/10
D2: ∫∫y^3 dxdy where 1 < x < 2y-1 and 1 < y < 2; I integrate and get 49/10.

Adding the two areas I get 31/5 which is perfect because the answer is 147/20.

Now please excuse me while I go kill myself.
splitting the triangle up is unnecessary. rather, write the non-horizontal lines in terms of y and integrate with respect to x first. the integral is given by:

$\int_1^2 \int_{2 - y}^{2y - 1}y^3~dx~dy$

any questions?

3. Don't worry, I have a really easy approach for you. Using this, your double integrals will become funny and nonsense, instead of troubling you.

Ok, forget everything, clear your mind, have fresh air..

Now, we have a double integral over a region, which is shown as $\int\int_D f(x,y)~dA$. You probably already know that $dA$ can be $dx~dy$ or $dy~dx$, depending on your solution.

Your region is a triangle, which is not a rectangle. Integrals on rectangular regions can be expressed as $\int_c^d\int_a^b f(x,y)~dy~dx$ or $\int_a^b \int_c^d f(x,y)~dx~dy$ where a b c d are constant and limits of the rectangle.

Since your region is not a rectangle, your integral can't be expressed using only constants like a b c d. We need functions here.

In order to use my approach, the region has to be bounded by two lines parallel to one of the axes. For example, if a region is bounded by "x=a and x=b" or "y=a and y=b", then we can go on. If it's not bounded, you should split the region. You may also need to split the region if the bottom and top functions are piecewise.

When we use $x=a$ and $x=b$ as bounds, the top and bottom functions are top and bottom bounds of the region. Also dA = dy dx. If the bottom bound is $f(x)$ and top bound is $g(x)$, our integral is: $\int_a^b\int_{f(x)}^{g(x)} \text{integrand}~dy~dx$

When we use $y=a$ and $y=b$ as bounds, the top and bottom functions are left and right bounds of the region. Also dA = dx dy. If the bottom bound is $f(y)$ and top bound is $g(y)$, our integral is: $\int_a^b\int_{f(y)}^{g(y)} \text{integrand}~dx~dy$

Code:
  |
|
2.o-----------o
|\********/
| \****/
1.|   o
|
__|_____________
|   1   2   3
Aha! This triangle is bounded by two lines parallel to x axis: $y=1$ and $y=2$. It also has non-piecewise top and bottom functions.

(red: top, blue: bottom)

Code:
  |
|
2.o-----------o
|\********/
| \****/
1.|   o
|
__|_____________
|   1   2   3
Now, you find f(y) and g(y), top and bottom functions.

Then, the integral will be $\int_1^2\int_{f(y)}^{g(y)}y^3~dx~dy$

4. Originally Posted by wingless
Don't worry, I have a really easy approach for you. Using this, your double integrals will become funny and nonsense, instead of troubling you.

Ok, forget everything, clear your mind, have fresh air..

Now, we have a double integral over a region, which is shown as $\int\int_D f(x,y)~dA$. You probably already know that $dA$ can be $dx~dy$ or $dy~dx$, depending on your solution.

Your region is a triangle, which is not a rectangle. Integrals on rectangular regions can be expressed as $\int_c^d\int_a^b f(x,y)~dy~dx$ or $\int_a^b \int_c^d f(x,y)~dx~dy$ where a b c d are constant and limits of the rectangle.

Since your region is not a rectangle, your integral can't be expressed using only constants like a b c d. We need functions here.

In order to use my approach, the region has to be bounded by two lines parallel to one of the axes. For example, if a region is bounded by "x=a and x=b" or "y=a and y=b", then we can go on. If it's not bounded, you should split the region. You may also need to split the region if the bottom and top functions are piecewise.

When we use $x=a$ and $x=b$ as bounds, the top and bottom functions are top and bottom bounds of the region. Also dA = dy dx. If the bottom bound is $f(x)$ and top bound is $g(x)$, our integral is: $\int_a^b\int_{f(x)}^{g(x)} \text{integrand}~dy~dx$

When we use $y=a$ and $y=b$ as bounds, the top and bottom functions are left and right bounds of the region. Also dA = dx dy. If the bottom bound is $f(y)$ and top bound is $g(y)$, our integral is: $\int_a^b\int_{f(y)}^{g(y)} \text{integrand}~dx~dy$

Code:
  |
|
2.o-----------o
|\********/
| \****/
1.|   o
|
__|_____________
|   1   2   3
Aha! This triangle is bounded by two lines parallel to x axis: $y=1$ and $y=2$. It also has non-piecewise top and bottom functions.

(red: top, blue: bottom)

Code:
  |
|
2.o-----------o
|\********/
| \****/
1.|   o
|
__|_____________
|   1   2   3
Now, you find f(y) and g(y), top and bottom functions.

Then, the integral will be $\int_1^2\int_{f(y)}^{g(y)}y^3~dx~dy$
"When we use and as bounds, the top and bottom functions are top and bottom bounds of the region. Also dA = dy dx."

I think that's where I went wrong! This was crystal clear and I appreciate it so much. I've been struggling a lot in this class for the past 2 weeks and the math tutors at school seriously make no sense. They're geniuses but all I hear coming out of their mouths are infinite this and that...

Thanks for the picture and spending your time to explain this stuff to me. The explanation of the bounds was very good and I understand it much better now.

Jhevon, thanks for your help again =)

5. Originally Posted by wingless
Don't worry, I have a really easy approach for you. Using this, your double integrals will become funny and nonsense, instead of troubling you.

Ok, forget everything, clear your mind, have fresh air..

Now, we have a double integral over a region, which is shown as $\int\int_D f(x,y)~dA$. You probably already know that $dA$ can be $dx~dy$ or $dy~dx$, depending on your solution.

Your region is a triangle, which is not a rectangle. Integrals on rectangular regions can be expressed as $\int_c^d\int_a^b f(x,y)~dy~dx$ or $\int_a^b \int_c^d f(x,y)~dx~dy$ where a b c d are constant and limits of the rectangle.

Since your region is not a rectangle, your integral can't be expressed using only constants like a b c d. We need functions here.

In order to use my approach, the region has to be bounded by two lines parallel to one of the axes. For example, if a region is bounded by "x=a and x=b" or "y=a and y=b", then we can go on. If it's not bounded, you should split the region. You may also need to split the region if the bottom and top functions are piecewise.

When we use $x=a$ and $x=b$ as bounds, the top and bottom functions are top and bottom bounds of the region. Also dA = dy dx. If the bottom bound is $f(x)$ and top bound is $g(x)$, our integral is: $\int_a^b\int_{f(x)}^{g(x)} \text{integrand}~dy~dx$

When we use $y=a$ and $y=b$ as bounds, the top and bottom functions are left and right bounds of the region. Also dA = dx dy. If the bottom bound is $f(y)$ and top bound is $g(y)$, our integral is: $\int_a^b\int_{f(y)}^{g(y)} \text{integrand}~dx~dy$

Code:
  |
|
2.o-----------o
|\********/
| \****/
1.|   o
|
__|_____________
|   1   2   3
Aha! This triangle is bounded by two lines parallel to x axis: $y=1$ and $y=2$. It also has non-piecewise top and bottom functions.

(red: top, blue: bottom)

Code:
  |
|
2.o-----------o
|\********/
| \****/
1.|   o
|
__|_____________
|   1   2   3
Now, you find f(y) and g(y), top and bottom functions.

Then, the integral will be $\int_1^2\int_{f(y)}^{g(y)}y^3~dx~dy$
Great Explanation!

--Chris