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Sin(x2 y2 da r , where r is the region in the first quadrant between the circles with center the origin and radii 1 and 5

Rudik [331]

I assume there's a plus sign missing above...

Convert to polar coordinates, using

$\begin{cases}x(r,\theta)=r\cos\theta\\y(r,\theta)=r\sin\theta\end{cases}$

Then the Jacobian is

$\dfrac{\partial(x,y)}{\partial(r,\theta)}=\begin{vmatrix}x_r&y_r\\x_\theta&y_\theta\end{vmatrix}=\begin{vmatrix}\cos\theta&\sin\theta\\r\sin\theta&-r\cos\theta\end{vmatrix}=-r$

Then

$\mathrm dA=\mathrm dx\,\mathrm dy=|-r|\,\mathrm dr\,\mathrm d\theta=r\,\mathrm dr\,\mathrm d\theta$

so the integral can be written as

$\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA=\int_0^{\pi/2}\int_1^5r\sin(r^2)\,\mathrm dr\,\mathrm d\theta$

Let $s=r^2$ , so that $\dfrac{\mathrm ds}2=r\,\mathrm dr$ .

$\displaystyle\frac12\int_0^{\pi/2}\int_1^{25}\sin s\,\mathrm ds\,\mathrm d\theta=-\frac12(\cos25-\cos1)\int_0^{\pi/2}\mathrm d\theta=\frac\pi4(\cos1-\cos25)$

3 0

3 years ago

Evaluate the following expression. Round your answer two decimal places. Log3(10)

Elanso [62]

$\bf \textit{Logarithm Change of Base Rule}
\\\\
log_a b\implies \cfrac{log_c b}{log_c a}\\\\
-------------------------------\\\\
log_3(10)\implies \cfrac{log_{10}(10)}{log_{10}(3)}\implies \cfrac{1}{log(3)}$

and you can just plug that in your calculator.

3 0

3 years ago

Read 2 more answers

Please help with the attached question. Thanks

Mademuasel [1]

Answer:

Choice A) $F(x) = 3\sqrt{x + 1}$ .

Step-by-step explanation:

What are the changes that would bring $G(x)$ to $F(x)$ ?

Translate $G(x)$ to the left by $1$ unit, and
Stretch $G(x)$ vertically (by a factor greater than $1$ .)

$G(x) = \sqrt{x}$ . The choices of $F(x)$ listed here are related to $G(x)$ :

Choice A) $F(x) = 3\;G(x+1)$ ;
Choice B) $F(x) = 3\;G(x-1)$ ;
Choice C) $F(x) = -3\;G(x+1)$ ;
Choice D) $F(x) = -3\;G(x-1)$ .

The expression in the braces (for example $x$ as in $G(x)$ ) is the independent variable.

To shift a function on a cartesian plane to the left by $a$ units, add $a$ to its independent variable. Think about how $(x-a)$ , which is to the left of $x$ , will yield the same function value.

Conversely, to shift a function on a cartesian plane to the right by $a$ units, subtract $a$ from its independent variable.

For example, $G(x+1)$ is $1$ unit to the left of $G(x)$ . Conversely, $G(x-1)$ is $1$ unit to the right of $G(x)$ . The new function is to the left of $G(x)$ . Meaning that $F(x)$ should should add $1$ to (rather than subtract $1$ from) the independent variable of $G(x)$ . That rules out choice B) and D).

Multiplying a function by a number that is greater than one will stretch its graph vertically.
Multiplying a function by a number that is between zero and one will compress its graph vertically.
Multiplying a function by a number that is between $-1$ and zero will flip its graph about the $x$ -axis. Doing so will also compress the graph vertically.
Multiplying a function by a number that is less than $-1$ will flip its graph about the $x$ -axis. Doing so will also stretch the graph vertically.

The graph of $G(x)$ is stretched vertically. However, similarly to the graph of this graph $G(x)$ , the graph of $F(x)$ increases as $x$ increases. In other words, the graph of $G(x)$ isn't flipped about the $x$ -axis. $G(x)$ should have been multiplied by a number that is greater than one. That rules out choice C) and D).

Overall, only choice A) meets the requirements.

Since the plot in the question also came with a couple of gridlines, see if the points $(x, y)$ 's that are on the graph of $F(x)$ fit into the expression $y = F(x) = 3\sqrt{x - 1}$ .