Answer:
8
Step-by-step explanation:
Hello,
[tex]x=3y^2<=>y=\sqrt{\dfrac{x}{3}} \ \ for \ x\geq 0[/tex]
And for y = 2, x = 3 * 2 * 2 = 12 so first, let's compute
[tex]\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[/tex]
The area which is asked is 12*2 - 16 = 24 - 16 = 8
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
Using integrals, it is found that the area of the region enclosed by the curves in the interval is of 27 units squared.
In this problem:
Then, the integral for the area is:
[tex]A = \int_{0}^{2} 3y^2 dy[/tex]
[tex]A = y^3|_{y = 0}^{y = 3}[/tex]
[tex]A = 3^3 - 0^3[/tex]
[tex]A = 27[/tex]
The area of the region enclosed by the curves in the interval is of 27 units squared.
You can learn more about the use of integrals to calculate an area at https://brainly.com/question/15127807
