Integration, Area under curve

Question

The question goes by : Find the finite region bounded by the curve

$$x=5t^2$$ $$y=2t^3$$ and the line $x = 5$. Find also the volume of the solid formed when this region is rotated through $\pi$ radians about the $x$ axis.

I got $4$ for the first question but the answer is $8$.

After integrating, I got $20[1/5t^5]$ with limits $1$ and $0$

Answer 1

You got the correct area in the first quadrant, bounded by the graph of the curve above and the $x$-axis below. If $t$ runs from $0$ to $1$, this is exactly the part of the curve that you get but that part of the curve alone together with the line $x=5$, does not bound a region.

However, if you allow negative values of $t$, you get the mirror image of that part of the curve w.r.t. the $x$-axis and the area enclosed by the curve and the line $x=5$ is double of what you found.

Since they do not specify an interval for the parameter $t$ in

The question goes by : Find the finite region bounded by the curve

$$x=5t^2$$ $$y=2t^3$$

you can probably assume they want you to consider the complete curve (let $t$ run through $\mathbb{R}$); only then you get a region bounded by the curve and the line $x=5$.