Math / Calculus

QuestionProblem 4 [30 points] Evaluate the integral by making an appropriate change of variables: $\iint_{R}\left(\frac{x-y}{x+y}\right)^{4} d A$ , where $R$ is the region bounded by the lines $x+y=1, x=0$ and $y=0$ .

Studdy Solution

STEP 1

What is this asking? Find the value of a double integral over a specific region by changing variables to make it easier. Watch out! Don't forget to adjust the limits of integration when changing variables!

STEP 2

1. Identify the region of integration
2. Choose a suitable change of variables
3. Transform the integral
4. Evaluate the transformed integral

STEP 3

Alright, let's first visualize the region $R$ .
It's bounded by the lines $x+y=1$ , $x=0$ , and $y=0$ .
This forms a right triangle with vertices at $(0,0)$ , $(1,0)$ , and $(0,1)$ .

STEP 4

To simplify the expression $\left(\frac{x-y}{x+y}\right)^{4}$ , let's introduce new variables: set $u = x+y$ and $v = x-y$ .
This way, our integrand becomes $\left(\frac{v}{u}\right)^{4}$ .

STEP 5

Next, we need the Jacobian of the transformation.
The relationships are: $x = \frac{u+v}{2}, \quad y = \frac{u-v}{2}$
Calculate the Jacobian determinant: $\frac{\partial(x, y)}{\partial(u, v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{vmatrix} = -\frac{1}{2}$

STEP 6

Now, let's rewrite the region $R$ in terms of $u$ and $v$ .
The line $x+y=1$ becomes $u=1$ , and the axes $x=0$ and $y=0$ imply $u=v$ and $u=-v$ , respectively.
So, the new region is bounded by $u=1$ , $v=u$ , and $v=-u$ .

STEP 7

The integral in new variables is: $\iint_{R'} \left(\frac{v}{u}\right)^{4} \left|-\frac{1}{2}\right| \, du \, dv$

STEP 8

Set up the limits: Since $u$ ranges from $0$ to $1$ and for each $u$ , $v$ ranges from $-u$ to $u$ , the integral becomes: $\int_{0}^{1} \int_{-u}^{u} \left(\frac{v}{u}\right)^{4} \cdot \frac{1}{2} \, dv \, du$

STEP 9

Evaluate the inner integral: $\int_{-u}^{u} \left(\frac{v}{u}\right)^{4} \, dv = \frac{1}{u^4} \int_{-u}^{u} v^4 \, dv$
The integral $\int_{-u}^{u} v^4 \, dv$ evaluates to: $\left[ \frac{v^5}{5} \right]_{-u}^{u} = \frac{u^5}{5} - \left(-\frac{u^5}{5}\right) = \frac{2u^5}{5}$

STEP 10

Substitute back into the outer integral: $\frac{1}{2} \int_{0}^{1} \frac{2u^5}{5u^4} \, du = \frac{1}{5} \int_{0}^{1} u \, du$

STEP 11

Evaluate the final integral: $\frac{1}{5} \left[ \frac{u^2}{2} \right]_{0}^{1} = \frac{1}{5} \cdot \frac{1}{2} = \frac{1}{10}$

STEP 12

The value of the integral is $\frac{1}{10}$ .

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QuestionProblem 4 [30 points] Evaluate the integral by making an appropriate change of variables: $\iint_{R}\left(\frac{x-y}{x+y}\right)^{4} d A$ , where $R$ is the region bounded by the lines $x+y=1, x=0$ and $y=0$ .

Studdy Solution

STEP 1

What is this asking? Find the value of a double integral over a specific region by changing variables to make it easier. Watch out! Don't forget to adjust the limits of integration when changing variables!

STEP 2

1. Identify the region of integration
2. Choose a suitable change of variables
3. Transform the integral
4. Evaluate the transformed integral

STEP 3

Alright, let's first visualize the region $R$ .
It's bounded by the lines $x+y=1$ , $x=0$ , and $y=0$ .
This forms a right triangle with vertices at $(0,0)$ , $(1,0)$ , and $(0,1)$ .

STEP 4

To simplify the expression $\left(\frac{x-y}{x+y}\right)^{4}$ , let's introduce new variables: set $u = x+y$ and $v = x-y$ .
This way, our integrand becomes $\left(\frac{v}{u}\right)^{4}$ .

STEP 5

STEP 6

Now, let's rewrite the region $R$ in terms of $u$ and $v$ .
The line $x+y=1$ becomes $u=1$ , and the axes $x=0$ and $y=0$ imply $u=v$ and $u=-v$ , respectively.
So, the new region is bounded by $u=1$ , $v=u$ , and $v=-u$ .

STEP 7

The integral in new variables is: $\iint_{R'} \left(\frac{v}{u}\right)^{4} \left|-\frac{1}{2}\right| \, du \, dv$

STEP 8

Set up the limits: Since $u$ ranges from $0$ to $1$ and for each $u$ , $v$ ranges from $-u$ to $u$ , the integral becomes: $\int_{0}^{1} \int_{-u}^{u} \left(\frac{v}{u}\right)^{4} \cdot \frac{1}{2} \, dv \, du$

STEP 9

Evaluate the inner integral: $\int_{-u}^{u} \left(\frac{v}{u}\right)^{4} \, dv = \frac{1}{u^4} \int_{-u}^{u} v^4 \, dv$
The integral $\int_{-u}^{u} v^4 \, dv$ evaluates to: $\left[ \frac{v^5}{5} \right]_{-u}^{u} = \frac{u^5}{5} - \left(-\frac{u^5}{5}\right) = \frac{2u^5}{5}$

STEP 10

Substitute back into the outer integral: $\frac{1}{2} \int_{0}^{1} \frac{2u^5}{5u^4} \, du = \frac{1}{5} \int_{0}^{1} u \, du$

STEP 11

Evaluate the final integral: $\frac{1}{5} \left[ \frac{u^2}{2} \right]_{0}^{1} = \frac{1}{5} \cdot \frac{1}{2} = \frac{1}{10}$

STEP 12

The value of the integral is $\frac{1}{10}$ .

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QuestionProblem 4 [30 points] Evaluate the integral by making an appropriate change of variables: ∬R(x−yx+y)4dA\iint_{R}\left(\frac{x-y}{x+y}\right)^{4} d A∬R​(x+yx−y​)4dA, where RRR is the region bounded by the lines x+y=1,x=0x+y=1, x=0x+y=1,x=0 and y=0y=0y=0.

Studdy Solution

STEP 1

What is this asking? Find the value of a double integral over a specific region by changing variables to make it easier. Watch out! Don't forget to adjust the limits of integration when changing variables!

STEP 2

1. Identify the region of integration2. Choose a suitable change of variables3. Transform the integral4. Evaluate the transformed integral

STEP 3

Alright, let's first **visualize the region** R R R.It's bounded by the lines x+y=1 x+y=1 x+y=1, x=0 x=0 x=0, and y=0 y=0 y=0.This forms a right triangle with vertices at (0,0)(0,0)(0,0), (1,0)(1,0)(1,0), and (0,1)(0,1)(0,1).

STEP 4

To simplify the expression (x−yx+y)4\left(\frac{x-y}{x+y}\right)^{4}(x+yx−y​)4, let's **introduce new variables**: set u=x+y u = x+y u=x+y and v=x−y v = x-y v=x−y.This way, our integrand becomes (vu)4\left(\frac{v}{u}\right)^{4}(uv​)4.

STEP 5

STEP 6

Now, let's **rewrite the region** R R R in terms of u u u and v v v.The line x+y=1 x+y=1 x+y=1 becomes u=1 u=1 u=1, and the axes x=0 x=0 x=0 and y=0 y=0 y=0 imply u=v u=v u=v and u=−v u=-v u=−v, respectively.So, the new region is bounded by u=1 u=1 u=1, v=u v=u v=u, and v=−u v=-u v=−u.

STEP 7

The **integral in new variables** is: ∬R′(vu)4∣−12∣ du dv\iint_{R'} \left(\frac{v}{u}\right)^{4} \left|-\frac{1}{2}\right| \, du \, dv∬R′​(uv​)4∣∣​−21​∣∣​dudv

STEP 8

**Set up the limits**: Since u u u ranges from 0 0 0 to 1 1 1 and for each u u u, v v v ranges from −u-u−u to u u u, the integral becomes: ∫01∫−uu(vu)4⋅12 dv du\int_{0}^{1} \int_{-u}^{u} \left(\frac{v}{u}\right)^{4} \cdot \frac{1}{2} \, dv \, du∫01​∫−uu​(uv​)4⋅21​dvdu

STEP 9

STEP 10

**Substitute back** into the outer integral: 12∫012u55u4 du=15∫01u du\frac{1}{2} \int_{0}^{1} \frac{2u^5}{5u^4} \, du = \frac{1}{5} \int_{0}^{1} u \, du21​∫01​5u42u5​du=51​∫01​udu

STEP 11

**Evaluate the final integral**: 15[u22]01=15⋅12=110\frac{1}{5} \left[ \frac{u^2}{2} \right]_{0}^{1} = \frac{1}{5} \cdot \frac{1}{2} = \frac{1}{10}51​[2u2​]01​=51​⋅21​=101​

STEP 12

The value of the integral is 110\frac{1}{10}101​.

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QuestionProblem 4 [30 points] Evaluate the integral by making an appropriate change of variables: $\iint_{R}\left(\frac{x-y}{x+y}\right)^{4} d A$ , where $R$ is the region bounded by the lines $x+y=1, x=0$ and $y=0$ .

1. Identify the region of integration
2. Choose a suitable change of variables
3. Transform the integral
4. Evaluate the transformed integral

Alright, let's first visualize the region $R$ .
It's bounded by the lines $x+y=1$ , $x=0$ , and $y=0$ .
This forms a right triangle with vertices at $(0,0)$ , $(1,0)$ , and $(0,1)$ .

To simplify the expression $\left(\frac{x-y}{x+y}\right)^{4}$ , let's introduce new variables: set $u = x+y$ and $v = x-y$ .
This way, our integrand becomes $\left(\frac{v}{u}\right)^{4}$ .

Now, let's rewrite the region $R$ in terms of $u$ and $v$ .
The line $x+y=1$ becomes $u=1$ , and the axes $x=0$ and $y=0$ imply $u=v$ and $u=-v$ , respectively.
So, the new region is bounded by $u=1$ , $v=u$ , and $v=-u$ .

The integral in new variables is: $\iint_{R'} \left(\frac{v}{u}\right)^{4} \left|-\frac{1}{2}\right| \, du \, dv$

Set up the limits: Since $u$ ranges from $0$ to $1$ and for each $u$ , $v$ ranges from $-u$ to $u$ , the integral becomes: $\int_{0}^{1} \int_{-u}^{u} \left(\frac{v}{u}\right)^{4} \cdot \frac{1}{2} \, dv \, du$

Substitute back into the outer integral: $\frac{1}{2} \int_{0}^{1} \frac{2u^5}{5u^4} \, du = \frac{1}{5} \int_{0}^{1} u \, du$

Evaluate the final integral: $\frac{1}{5} \left[ \frac{u^2}{2} \right]_{0}^{1} = \frac{1}{5} \cdot \frac{1}{2} = \frac{1}{10}$

The value of the integral is $\frac{1}{10}$ .