Math  /  Calculus

QuestionMath 263 Exam 3A November 8, 2024
6. (20 points) Evaluate 01/2y21y224xy24y31+(x+y2)3dxdy\int_{0}^{\sqrt{1 / 2}} \int_{y^{2}}^{1-y^{2}} \frac{24 x y-24 y^{3}}{1+\left(x+y^{2}\right)^{3}} d x d y, by using the change of variables u=xy2u=x-y^{2} and v=x+y2v=x+y^{2}. Hint: J(x,y)=(u,v)(x,y)J(x, y)=\left|\frac{\partial(u, v)}{\partial(x, y)}\right| x=1y2x=1-y^{2}

Studdy Solution

STEP 1

What is this asking? We need to compute a double integral using a suggested change of variables. Watch out! Don't forget to calculate the Jacobian and change the limits of integration!

STEP 2

1. Define the change of variables and compute the Jacobian.
2. Transform the integrand.
3. Determine the new limits of integration.
4. Evaluate the transformed integral.

STEP 3

We're given the change of variables u=xy2u = x - y^2 and v=x+y2v = x + y^2.
Let's **compute the Jacobian**!
Remember, the Jacobian measures how much our area element changes when we switch coordinate systems.
It's super important for making sure our integral stays accurate after the transformation!

STEP 4

The Jacobian is given by J(x,y)=(u,v)(x,y)J(x, y) = \left| \frac{\partial(u, v)}{\partial(x, y)} \right|.
This means we need to find the absolute value of the determinant of the matrix of partial derivatives.

STEP 5

Let's **calculate those partial derivatives**: ux=1,uy=2y\frac{\partial u}{\partial x} = 1, \quad \frac{\partial u}{\partial y} = -2y vx=1,vy=2y\frac{\partial v}{\partial x} = 1, \quad \frac{\partial v}{\partial y} = 2y

STEP 6

Now, let's **form the matrix and compute the determinant**: J(x,y)=12y12y=4y=4yJ(x, y) = \begin{vmatrix} 1 & -2y \\ 1 & 2y \end{vmatrix} = |4y| = 4y Since 0y120 \le y \le \sqrt{\frac{1}{2}}, we know that yy is non-negative, so we can drop the absolute value signs.
Thus, J(x,y)=4yJ(x,y) = 4y.

STEP 7

We need J(u,v)J(u,v), which is 1J(x,y)\frac{1}{J(x,y)}.
So, J(u,v)=14yJ(u, v) = \frac{1}{4y}.
This tells us how to scale our integral after the transformation.

STEP 8

We have u=xy2u = x - y^2 and v=x+y2v = x + y^2.
Adding these equations gives u+v=2xu + v = 2x, so x=u+v2x = \frac{u + v}{2}.
Subtracting the equations gives vu=2y2v - u = 2y^2, so y2=vu2y^2 = \frac{v - u}{2}.

STEP 9

Now, let's **rewrite the integrand** in terms of uu and vv: 24xy24y31+(x+y2)3=24y(xy2)1+(x+y2)3=24yu1+v3\frac{24xy - 24y^3}{1 + (x + y^2)^3} = \frac{24y(x - y^2)}{1 + (x + y^2)^3} = \frac{24y \cdot u}{1 + v^3}

STEP 10

The original limits are y2x1y2y^2 \le x \le 1 - y^2 and 0y120 \le y \le \sqrt{\frac{1}{2}}.

STEP 11

Since u=xy2u = x - y^2, the lower limit for xx becomes u=y2y2=0u = y^2 - y^2 = 0, and the upper limit becomes u=(1y2)y2=12y2u = (1 - y^2) - y^2 = 1 - 2y^2.

STEP 12

Since v=x+y2v = x + y^2, we have v=y2+y2=2y2v = y^2 + y^2 = 2y^2 and v=(1y2)+y2=1v = (1 - y^2) + y^2 = 1.

STEP 13

Since v=2y2v = 2y^2, we have y=v2y = \sqrt{\frac{v}{2}}.
Also, u=12y2=1vu = 1 - 2y^2 = 1 - v.
The limits for yy become 0v2120 \le \sqrt{\frac{v}{2}} \le \sqrt{\frac{1}{2}}, which simplifies to 0v10 \le v \le 1.

STEP 14

Our transformed integral is: 0101v24v2u1+v314v2dudv=0101v6u1+v3dudv\int_{0}^{1} \int_{0}^{1-v} \frac{24 \sqrt{\frac{v}{2}} \cdot u}{1 + v^3} \frac{1}{4\sqrt{\frac{v}{2}}} du dv = \int_{0}^{1} \int_{0}^{1-v} \frac{6u}{1 + v^3} du dv

STEP 15

**Integrating with respect to** uu: 01[3u21+v3]01vdv=013(1v)21+v3dv=013(12v+v2)1+v3dv\int_{0}^{1} \left[ \frac{3u^2}{1 + v^3} \right]_{0}^{1-v} dv = \int_{0}^{1} \frac{3(1 - v)^2}{1 + v^3} dv = \int_{0}^{1} \frac{3(1 - 2v + v^2)}{1 + v^3} dv

STEP 16

Notice that 1+v3=(1+v)(1v+v2)1 + v^3 = (1+v)(1-v+v^2) and (1v)2=12v+v2(1-v)^2 = 1-2v+v^2.
If we divide 12v+v21-2v+v^2 by 1v+v21-v+v^2, we get 1v1v+v21-\frac{v}{1-v+v^2}.
This doesn't seem to help.
Let's try partial fractions.
We can factor the denominator as (1+v)(1v+v2)(1+v)(1-v+v^2).
This gives us 013(12v+v2)(1+v)(1v+v2)dv=0131+vdv=[3ln(1+v)]01=3ln(2)3ln(1)=3ln(2)\int_0^1 \frac{3(1-2v+v^2)}{(1+v)(1-v+v^2)} dv = \int_0^1 \frac{3}{1+v} dv = [3\ln(1+v)]_0^1 = 3\ln(2) - 3\ln(1) = 3\ln(2)

STEP 17

The value of the integral is 3ln(2)3\ln(2).

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