Solve $\int_{0}^{1}\int_{0}^{x}\int_{z}^{x}x\cos (y^{2})dydzdx$

Question

I want to solve this triple integral: $$\int_{0}^{1}\int_{0}^{x}\int_{z}^{x}x\cos (y^{2})dydzdx$$

I tried to change the variables but I still couldn't make it. Is there any clue how to start?

Answer 1

The variable $x$ can take any value from $0$ to $1$ and, for each value of $x$, $z$ can take any value from $0$ to $x$ and $y$ can take any value from $z$ to $x$. So, if you fix $x\in[0,1]$, $y$ can take any value from $0$ to $x$ and $z$ can take any value from $0$ to $y$. Therefore\begin{align}\int_0^1\int_0^x\int_z^xx\cos(y^2)\,\mathrm dy\,\mathrm dz\,\mathrm dx&=\int_0^1\int_0^x\int_0^yx\cos(y^2)\,\mathrm dz\,\mathrm dy\,\mathrm dx\\&=\int_0^1\int_0^xxy\cos(y^2)\,\mathrm dy\,\mathrm dx\\&=\int_0^1\frac12x\sin(x^2)\,\mathrm dx\\&=\frac14-\frac{\cos(1)}4.\end{align}

Answer 2

You need to determine which order of integration will make the calculations possible. Notice that the region of integration is defined by $0 \leq z \leq y \leq x \leq 1$.

As written, you need to first integrate with respect to $y$, but that will not work since you have $\cos(y^2)$ and no $y$ outside that.

What if you integrate first with respect to $x$? That is possible, and the integral becomes: $$\int_0^1 \int_z^1 \int_y^1 x\cos(y^2) \ dx \ dy \ dz \quad \text{or} \quad \int_0^1\int_0^y\int_y^1 x\cos(y^2) \ dx \ dz \ dy.$$ In either case the innermost integral is: $$\int_y^1 x \cos(y^2) \ dx = \frac{1}{2}x^2\cos(y^2)\Big|_{x=y}^{x=1}=\frac{1}{2}\cos(y^2)-\frac{1}{2}y^2\cos(y^2).$$ We can't integrate either of these with respect to $y$, so we would want to integrate with respect to $z$ next: \begin{align*}\int_0^y \left(\frac{1}{2}\cos(y^2)-\frac{1}{2}y^2\cos(y^2)\right) \ dz &=\left(\frac{1}{2}\cos(y^2)-\frac{1}{2}y^2\cos(y^2)\right)z\Big|_{x=0}^{x=y}\\ &= y\left(\frac{1}{2}\cos(y^2)-\frac{1}{2}y^2\cos(y^2)\right)\\ &=\frac{1}{2}\cos(y^2)(y-y^2). \end{align*} Again we can't integrate with respect to $y$.

So now we backtrack and instead try integrating with respect to $z$ first. The innermost integral is: $$\int_0^y x\cos(y^2) \ dz = xz\cos(y^2)\Big|_{z=0}^{z=y}=xy\cos(y^2).$$ We could integrate with respect to $x$ or $y$ next. But notice that integrating with respect to $x$ will introduce a $\cos(y^2)$ and a $y^2\cos(y^2)$ term (since the bounds on $x$ are $y \leq x \leq 1$, and the antiderivative of $x$ is $\frac{1}{2}x^2$). So we should integrate with respect to $y$ next, which we can do using a $u$-substitution. After that, it turns out you can also do the final integral with respect to $x$. So we have determined that the best order of integration is: $$\int_0^1 \int_0^x \int_0^y x \cos(y^2) \ dz \ dy \ dx.$$

Altogether there are $3!=6$ possible orders of integration to consider, and you need to be systematic about checking the possibilities. We could immediately rule out two of them ($dy \ dx \ dz$ and $dy \ dz \ dx$) because we cannot do the innermost integral. For the other four, we had to do some trial and error, but it can be sped up by thinking ahead about the next step, as I did above when deciding whether to integrate $xy\cos(y^2)$ with respect to $x$ or $y$ first.