never_lose
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- Jul 9, 2011
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- 7
Evaluate the double integral.
\(\displaystyle \int \int_{\Omega} (4 - y^2) dx dy\)
\(\displaystyle \Omega\) being the bounded region between \(\displaystyle y^2 = 2x\) and \(\displaystyle y^2 = 8 - 2x\)
I'm not sure how to set this up. The book sets it up like this:
\(\displaystyle \int_{-2}^{2} \int_{\frac{1}{2}y^2}^{4-\frac{1}{2}y^2} (4 - y^2) dx dy\)
I can kinda see how they got the -2 to 2 for the y integral, because solving the system for x, you get x = 2.
Then solving the first equation for y you get:
\(\displaystyle y = \pm \sqrt {2x}\)
\(\displaystyle -\sqrt {2x} \le y \le \sqrt {2x}\)
\(\displaystyle -2 \le y \le 2\)
I have no idea how they got \(\displaystyle \int_{\frac{1}{2}y^2}^{4-\frac{1}{2}y^2}\), though.
\(\displaystyle \int \int_{\Omega} (4 - y^2) dx dy\)
\(\displaystyle \Omega\) being the bounded region between \(\displaystyle y^2 = 2x\) and \(\displaystyle y^2 = 8 - 2x\)
I'm not sure how to set this up. The book sets it up like this:
\(\displaystyle \int_{-2}^{2} \int_{\frac{1}{2}y^2}^{4-\frac{1}{2}y^2} (4 - y^2) dx dy\)
I can kinda see how they got the -2 to 2 for the y integral, because solving the system for x, you get x = 2.
Then solving the first equation for y you get:
\(\displaystyle y = \pm \sqrt {2x}\)
\(\displaystyle -\sqrt {2x} \le y \le \sqrt {2x}\)
\(\displaystyle -2 \le y \le 2\)
I have no idea how they got \(\displaystyle \int_{\frac{1}{2}y^2}^{4-\frac{1}{2}y^2}\), though.
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