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Solve the area of the shaded region.

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Answered 1 month ago
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Answered 1 month ago
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As we can see the points of intersection are (3,3)(-3, 3) and (0,0)(0, 0). The area of the region sought is best to evaluate by integrating with respect to yy since the equations are expressed as xx in terms of yy. We also can rewrite them xR=2yy2x_R = 2y-y^2 (because this curve is on the right) and xL=y24yx_L = y^2 - 4y (because this curve is on the left). So a typical rectangle will have a length of xRxL=2yy2(y24y)=6y2y2x_R - x_L = 2y-y^2 - (y^2 - 4y) = 6y - 2y^2 and width of Δy\Delta y (or dydy in integration). Therefore, the area of the region bounded by the two curves is

A=03(6y2y2)dy=[3y22y33]03=(3(3)32(3)33)(3(0)32(0)33)=90=9\begin{align*} A &=\int_{0}^{3} \left(6y - 2y^2\right) dy\\ &=\left[ 3y^2 - \dfrac{2y^3}{3}\right]_{0}^{3}\\ &=\left(3(3)^3 - \dfrac{2(3)^3}{3} \right) -\left(3(0)^3 - \dfrac{2(0)^3}{3} \right)\\ &=9 - 0\\ &=9 \end{align*}

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