Question #119357

Let D={(x,y)∈R^2:y^2<x<y and 0<y<1}. The answer to ∬D 2y^2xdA


is

Select one:

a. 70

b. 1/35

c. -70

d. 2/35

e. 35

Expert's answer

Since region bounded by

x:y2→y,        y:0→1x:y^2\to y,\ \ \ \ \ \ \ \ y :0\to 1

Then

∫∫Dy2xdA=∫01∫y2yy2xdxdy=12∫01y2x2∣y2ydy=12∫01y2(y2−y4)dy=12∫01(y4−y6)dy=12(15y5−17y7)∣01=12(15−17)=135\begin{aligned} \int\int_{D} y^2xdA&= \int_{0}^{1}\int_{y^2}^{y} y^2xdxdy \\ &= \frac{1}{2}\int_{0}^{1} y^2x^2\bigg|_{y^2}^{y} dy \\ &=\frac{1}{2}\int_{0}^{1} y^2(y^2-y^4)dy\\ &= \frac{1}{2}\int_{0}^{1} (y^4-y^6)dy\\ &= \frac{1}{2} (\frac{1}{5}y^5-\frac{1}{7}y^7)\bigg|_{0}^{1}\\ &= \frac{1}{2}( \frac{1}{5} -\frac{1}{7})\\ &= \frac{1}{35} \end{aligned}


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