My Homework - Can You Help Me?

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  • #1
ozgunozgur
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I am trying to solve these questions for hours :/

fffff.png
 
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  • #2
MarkFL
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Let's begin with the first problem...what have you tried so far?
 
  • #3
ozgunozgur
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Let's begin with the first problem...what have you tried so far?
 

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  • #4
MarkFL
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Yes, I agree with your finding of a minimum at \((2,-1)\). It is a global minimum. For the second problem, isn't the integrand:

\(\displaystyle e^{y^8}\) ?
 
  • #5
ozgunozgur
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Yes, I agree with your finding of a minimum at \((2,-1)\). It is a global minimum. For the second problem, isn't the integrand:

\(\displaystyle e^{y^8}\) ?
Ah sorry. My third question is true? And second is a bit problem.
 
  • #6
ozgunozgur
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Yes, I agree with your finding of a minimum at \((2,-1)\). It is a global minimum. For the second problem, isn't the integrand:

\(\displaystyle e^{y^8}\) ?
Isn't it gamma function for 8y?
 
  • #7
ozgunozgur
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I did question 2. Is it true? Can you help me?
 

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  • #8
ozgunozgur
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Please help me for second question, I guess I have to draw a sketch. Can you draw it full?
 
  • #9
ozgunozgur
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Can you review my answer?
 

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  • #12
MarkFL
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The use of exponential integral and gamma functions goes beyond what I was taught in Calc 2. We did multi-variable calculus in Calc 3, but still worked with elementary functions. I can't help you with this problem. Perhaps someone else can.
 
  • #13
I like Serena
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Questions 1 and 3 are fairly straight forward applications of multivariable Calculus with elementary functions.
So I think question 2 must also be such a straight forward application.
Looks to me as if the question should read:
$$\int_0^1 \int_{\sqrt x}^1 \exp(y^3)\,dy\,dx=\,?$$
That is, with power $3$, and with the variables of integration swapped.
Now we can solve it by swapping the order of integration, which is likely intended. And yes, a graph may help.
 
  • #14
ozgunozgur
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Questions 1 and 3 are fairly straight forward applications of multivariable Calculus with elementary functions.
So I think question 2 must also be such a straight forward application.
Looks to me as if the question should read:
$$\int_0^1 \int_{\sqrt x}^1 \exp(y^3)\,dy\,dx=\,?$$
That is, with power $3$, and with the variables of integration swapped.
Now we can solve it by swapping the order of integration, which is likely intended. And yes, a graph may help.
I'm not skilled in this part. Please help continue.
 
  • #15
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I'm not skilled in this part. Please help continue.
Please show an attempt to swap the order of integration.
Or otherwise give us a clue in some detail where you are stuck.
You should have an example in your textbook that shows how it is done.
If you can't find such an example, you might take a look at this example.
 
  • #16
ozgunozgur
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Please show an attempt to swap the order of integration.
Or otherwise give us a clue in some detail where you are stuck.
You should have an example in your textbook that shows how it is done.
If you can't find such an example, you might take a look at this example.
 

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  • #17
MarkFL
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Now that Klaas has figured out what #2 is actually supposed to be, let's look first at the region over which we are to integrate:

mhb_0017.png


Reversing the order of integration, we may write:

\(\displaystyle I=\int_0^1\int_0^{y^2} e^{y^3}\,dx\,dy\)

Can you proceed?
 
  • #18
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For question 3 we should use a switch to polar coordinates:
\begin{tikzpicture}
\filldraw[green!70,draw=gray] (0,0) circle (2.5);
\draw[help lines] (-3.2,-3.2) grid (3.2,3.2);
\draw[->] (-3.4,0) -- (3.4,0);
\draw[->] (0,-3.2) -- (0,3.2);
\draw foreach \i/\x in {-2.5/-a,2.5/a} { (\i,0) node[below] {$\x$} };
\draw foreach \i/\y in {-2.5/-a,2.5/a} { (0,\i) node[ left ] {$\y$} };
\end{tikzpicture}
\[ \iint_D (x^2+y^2)\,dA = \int_0^a \int_0^{2\pi} r^2 \cdot r\,d\theta \,dr \]
 
  • #19
ozgunozgur
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Thank you very much.
https://ibb.co/0XxGrDr
https://ibb.co/ZKNtgZy
I rearranged I wonder if I wrote extra or are we going right?
 
  • #20
ozgunozgur
27
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.
 
  • #21
ozgunozgur
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I was sent the other solut. Can you edit these two solving. https://ibb.co/c6S7r8n
 
  • #22
ozgunozgur
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