Restricted Area and Integration
- Thread starter sunrise
- Start date
D
Deleted member 4993
Guest
What are the boundaries of R?
y goes from 0 to x1/3 and x goes from 0 to π.
Now continue...
Please share your work with us.
You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:
http://www.freemathhelp.com/forum/th...217#post322217
We can help - we only help after you have shown your work - or ask a specific question (e.g. "are these correct?")
^ My fault, I took too long searching for an online graphing program 
After drawing your region, decide numerical bounds on one variable, say y. Then the bounds on x will be functions of y.
From the picture you can see \(\displaystyle 0\le y\le \sqrt[3]{\pi}\) and \(\displaystyle y^3\le x \le \pi\).
\(\displaystyle \displaystyle \int_R y^3 \cos(x) dA =\int_0^{\sqrt[3]{\pi}}\left(\int_{y^3}^{\pi} y^3 \cos(x)dx\right)dy\)
After drawing your region, decide numerical bounds on one variable, say y. Then the bounds on x will be functions of y.
From the picture you can see \(\displaystyle 0\le y\le \sqrt[3]{\pi}\) and \(\displaystyle y^3\le x \le \pi\).
\(\displaystyle \displaystyle \int_R y^3 \cos(x) dA =\int_0^{\sqrt[3]{\pi}}\left(\int_{y^3}^{\pi} y^3 \cos(x)dx\right)dy\)
Last edited:
There is nothing wrong with the question. If you show your work, we may be able to pinpoint where an error may be occurring. Assuming your calculus is fine, my first blind guess (having no knowledge of your work) would be that your calculator may not be set to radian mode.