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Find the area of the region bounded by the graphs of the given functions x=3y²−9 x=0 y=0 y=1

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Answer to a math question Find the area of the region bounded by the graphs of the given functions x=3y²−9 x=0 y=0 y=1

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Clarabelle
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60 Answers
To find the area of the region bounded by the given functions, we need to first find the points of intersection.

Setting the two given functions equal to each other:

3y^2 - 9 = 0

y^2 = 3

y = \pm \sqrt{3}

So, the points of intersection are at (0,0) and at \left(\pm \sqrt{3}, \frac{1}{2}\right) .

Since we are interested in the area between x=3y^2-9 and x=0 , we need to integrate the difference of their equations with respect to y from y=0 to y=1 .

A = \int_{0}^{1} \left(3y^2-9-0\right) \,dy

A = \int_{0}^{1} (3y^2-9) \, dy

A = \left[y^3-9y\right]_{0}^{1}

A = \left(1^3-9(1)\right) - \left(0^3-9(0)\right)

A = 1-9 = -8

Therefore, the area of the region bounded by the graphs of the given functions is \boxed{8 sq. units} .

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