# The area of curve $y^3=x$ bounded by lines.

Find the area of region on $xy$ plane shaded by curve $y^3=x$ and lines $y=1$ and $x=8$

My solution: The line $x=8$ intersect curve at point $y=2$ and line $y=1$ intersect at point x=1. So the intended area is $$=\left(1-\int \limits_{0}^{1}y^3dy\right)+7=8-\frac{1}{4}=7,75$$

However, the answer in the book is $4,25$.

Can anyone point out where is my mistake with graphs? I have rechecked many times and did not detected it.

Referring to the graph:

1-method: $$S=\int_{x=1}^{x=8} x^{1/3}dx-7\cdot 1=\left(\frac{3}{4} x^{4/3}\right) \bigg|_1^8-7=\left(12-\frac34\right)-7=4\frac14.$$

2-method: $$S=8\cdot 1-\int_{y=1}^{y=2} y^3dy=8-\left(\frac{1}{4} y^4\right) \bigg|_1^2=8-\left(4-\frac14\right)=4\frac14.$$

• Good. But Let me ask you one question. The lower part of figure (under the line $y=1$) is also bounded by curve and two lines right? – ZFR Sep 27 '17 at 6:58
• No, if below $y=1$ is considered, then the figure will be bounded by the curve on the left, by the line $y=1$ above and by the line $x=8$ on the right, however, no bound from below. – farruhota Sep 27 '17 at 7:03
• So, the only region bounded by all three functions is the shaded region indicated on the graph above. – farruhota Sep 27 '17 at 7:04
• It will be bounded from below by $x$-axis right? Though there is nothing about that, right? – ZFR Sep 27 '17 at 7:06
• It depends on with respect to $x$ or $y$. If w.r.t. $x$, then bounded by $y=x^{1/3}$ from above and $y=1$ from below. If w.r.t. $y$, then bounded by $x=8$ from above and $x=y^3$ from below. So, look at the $\Delta x$ and $\Delta y$. – farruhota Sep 27 '17 at 7:47

I think you didn't understand well what area you should calculate. You should calculate the area above the $y=1$ line and hence it would be:

$$\int_1^2 (8 - y^3)dy = 8y - \frac {y^4}{4}\bigg|_1^8 = 16 - 8 -\frac{16}{4} + \frac 14 = 4 + \frac 14$$

You need to find the area of the shaded region.

• I have drawn a picture but I can't see anything false. Could you provide picture please? – ZFR Sep 27 '17 at 6:55
• @RFZ I've added one in the post – Stefan4024 Sep 27 '17 at 7:19