A flying disc has a radius of

Evaluate the integral e^xy w region d xy=1, xy=4, x/y=1, x/y=2

LUCKY_DIMON [66]

Make a change of coordinates:

$u(x,y)=xy$
$v(x,y)=\dfrac xy$

The Jacobian for this transformation is

$\mathbf J=\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial v}{\partial x}\\\\\dfrac{\partial u}{\partial y}&\dfrac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}y&x\\\\\dfrac1y&-\dfrac x{y^2}\end{bmatrix}$

and has a determinant of

$\det\mathbf J=-\dfrac{2x}y$

Note that we need to use the Jacobian in the other direction; that is, we've computed

$\mathbf J=\dfrac{\partial(u,v)}{\partial(x,y)}$

but we need the Jacobian determinant for the reverse transformation (from $(x,y)$ to $(u,v)$ . To do this, notice that

$\dfrac{\partial(x,y)}{\partial(u,v)}=\dfrac1{\dfrac{\partial(u,v)}{\partial(x,y)}}=\dfrac1{\mathbf J}$

we need to take the reciprocal of the Jacobian above.

The integral then changes to

$\displaystyle\iint_{\mathcal W_{(x,y)}}e^{xy}\,\mathrm dx\,\mathrm dy=\iint_{\mathcal W_{(u,v)}}\dfrac{e^u}{|\det\mathbf J|}\,\mathrm du\,\mathrm dv$
$=\displaystyle\frac12\int_{v=}^{v=}\int_{u=}^{u=}\frac{e^u}v\,\mathrm du\,\mathrm dv=\frac{(e^4-e)\ln2}2$

8 0

3 years ago

The smiths are building a ramp to their porch. The porch is 5 feet off the ground. Building regulations state that the ramp must

Annette [7]

I am pretty sure it is 40 feet

5 0

3 years ago

ANSWER THISSSSSSSSSS

lakkis [162]

Answer:

it depends on how its being answered so i'm going to answer its by what i think it is (20)

Step-by-step explanation:

technically is every piece of clothing in one outfit like one different sock to the same parts of an outfit is the same thing right?

i answered this the best i could im sorry if it wrong

8 0

3 years ago

AREA ADDITION AND SUBTRACTION URGENT?

Novay_Z [31]

The area of shaded region is

$A_{shaded}=A_{circle}-A_{segment}$ .

Find the area of circle and segment:

$A_{circle}=\pi r^2=\pi (8.35)^2=69.7225\pi=219.0397,\\ A_{segment}=A_{sector}-A_{triangle},\\ \\A_{sector}=\dfrac{r^2 \alpha}{2} =\dfrac{(8.35)^2\cdot \frac{5}{9}\pi}{2}=19.3674\pi=60.8445,\\ \\A_{triangle}=\dfrac{1}{2}r^2\sin \alpha=\dfrac{1}{2}(8.35)^2\sin 100^{\circ}=34.3316,\\ \\A_{segment}=60.8445-34.3316=26.5129$ .

Then $A_{shaded}=A_{circle}-A_{segment} =219.0397-26.5129=192.5268$ .

The area of shaded region rounded to the nearest tenth is 192.5 sq. ft.

Answer: 192.5 sq. ft.

6 0

3 years ago

What is the inverse of f(x) = 8x - 7

nignag [31]

Y=(x+7)/8

Work:
Y=8x-7 —> x=8y-7
Solve for y
x+7=8y
y=(x+7)/8

3 0

3 years ago