Mathematics, 12.08.2020 04:01 jpsaad00
Find the area of the region enclosed by the curves x=3y^2, x=0, and y=2
Answers: 3
, hence the integral is relative to y.The lower limit is when x = 0, hence 
.The upper limit is when y = 2.
![\displaystyle \int\limits^{12}_0 {\sqrt{\dfrac{x}{3}}} \, dx =\dfrac{1}{\sqrt{3}} \int\limits^{12}_0 {\sqrt{x}} \, dx\\\\=\dfrac{1}{\sqrt{3}} \left[ \dfrac{2}{3}x^{3/2}\right]_0^{12}\\\\=\dfrac{1}{\sqrt{3}} *\dfrac{2}{3}*12*\sqrt{12}\\\\=\dfrac{2*12*2*\sqrt{3}}{3*\sqrt{3}}\\\\=2*4*2=16](/tpl/images/0717/8271/04c99.png)
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Find the area of the region enclosed by the curves x=3y^2, x=0, and y=2...
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