How Do You Integrate y^3 Over a Triangle with Given Vertices?

[ineedhelpnow]

$\int \, \int_{D}^{} \, y^3 dA$ D is the triangular region with vertices (0,1), (1,2), (4,1)

i can't get past this problem. i drew the triangle but i don't know how to find the intervals...
excuse my ugly drawing :p
View attachment 3575

 

[I like Serena]

$\int \, \int_{D}^{} \, y^3 dA$ D is the triangular region with vertices (0,1), (1,2), (4,1)

i can't get past this problem. i drew the triangle but i don't know how to find the intervals...
excuse my ugly drawing :p
View attachment 3575

Hi! ;)

Let's start at $y=1$.
We can draw a thin horizontal rectangle there with length $4$ and height $dy$.

At $y=2$, we would have a rectangle with length $0$ and height $dy$.

Everywhere between, we have a linear transition.
That is, we have a rectangle of length $(8y-4)$ and height $dy$.

That totals to:
$$\iint_D y^3 dA = \int_1^2 y^3 \cdot (8y-4) \cdot dy$$

 

[ineedhelpnow]

hey ils.

uuum (Blush) its a double integral though.

 

[I like Serena]

hey ils.

uuum (Blush) its a double integral though.

Yep. (Nod)

So it's:
\begin{aligned}\iint_D y^3dA &= \int_1^2 \int_{\text{weird range of length $(8 - 4y)$}} y^3dx dy \\
&= \int_1^2 y^3 \left(\int_{\text{weird range of length $(8 - 4y)$}} dx\right) dy \\
&=\int_1^2 y^3 (8-4y)dy\end{aligned}
(Wasntme)

 

[ineedhelpnow]

(Rofl) how do i find the weird range? so like $D={(x,y)|?\le x \le ?, ? \le y \le ?}$. i was looking at an example on chegg and they used the equation from side AB and BC for the interval of x. i don't remember how to get the equation of the sides of a triangle (Blush)

 

[I like Serena]

(Rofl) how do i find the weird range? so like $D={(x,y)|?\le x \le ?, ? \le y \le ?}$. i was looking at an example on chegg and they used the equation from side AB and BC for the interval. i don't remember how to get the equation of the sides of a triangle (Blush)

What is the range of $x$ when $y=1$?
What if $y=2$?
And what if $y$ has some value in between those 2? (Wondering)We can also do it with the equations of AB and BC, but that is more complicated. (Doh)Anyway, for the equation of AB, we have the 2 points (0,1) and (1,2).
The equation for AB is given by:
$$y-y_A=\frac{y_B-y_A}{x_B-x_A}(x-x_A)$$
That is:
$$y-1=\frac{2-1}{1-0}(x-0) \quad\Rightarrow\quad y=x+1$$
(Wasntme)

 

[ineedhelpnow]

thank you ils
yeah the reason why i asked for the equation of the line is because this lesson is about double integrals where one of the integrals is between a range of two equations and the other is between two actual numbers.

 

[MarkFL]

Another formulation would be:

$$V=\iint\limits_{R}y^3\,dA=\int_1^2 y^3\int_{y-1}^{7-3y}\,dx\,dy=\int_1^2 y^3(8-4y)\,dy$$

Just as suggested above. :D

 

[ineedhelpnow]

i found BC and i got $x=3y-5$.
so $ \int_{y-1}^{3y-5} \ $?

 

[I like Serena]

i found BC and i got $x=3y-5$.
so $ \int_{y-1}^{3y-5} \ $?

That doesn't look right. (Worried)
How did you find it? (Wondering)

 

[ineedhelpnow]

i have no idea. i did it again. its x=7-3y