Finding the volume of the solid below $z=y$ and above the region in the $xy$-plane bounded by $y=x$ and $y=x^2$. (Seeking advice on bounds.)

cylindrical coordinatesintegrationspherical coordinates

So, I have this question which states:

Find the volume of the solid that lies below the surface,
$$z=y$$
and above the region $T$ in the $xy$-plane enclosed by the two curves $y=x$ and $y = x^2$.

enter image description here

As shown in the diagram.

And I realise this is through triple integrals and was wondering what the bounds would be.

My initial thoughts are let us use spherical coordinates: $$ 0 \leq \theta \leq 2 \pi$$
$$0 \leq \phi \leq \frac{\pi}{4}$$
and for $$\rho$$ I'm not too sure.

I was hoping someone could clarify what the bounds would be.
A simple (or complex) explanation would really help thank you

Best Answer

Yes you have to do triple integral. You can just do it in cartesian coordinates.

$0 \leq z \leq y, x^2 \leq y \leq x, 0 \leq x \leq 1$

Order of integration will be $dz$ first, then $\ dy$ and $\ dx$ last. If you change the order of integration, the limits will have to change accordingly.

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