In my PDE class, I have come across an integral in the form $$ \int_{B(x,t)} \psi(y) dy. $$
Then, the author makes the substitution $y = x + tw$ where $w = (r\cos\theta, r\sin\theta) \in B(0,1)$. Apparently, this implies that $dy = t^2 dw$ and the integral becomes
$$ \int_{B(0,1)} \psi(x + tw) t^2 dw. $$
I understand why the integral is now over $B(0,1)$, but I don't understand why $dy = t^2 dw$. Can someone explain?
Consider the function $\varphi(w)=x+tw$, where $x,t$ are fixed. Then, integrating over the image of $\varphi$ is the same as integrating over the ball of radius $t$ centered at $x$. Further, the Jacobian determinant is $t^2.$ Now, you just directly apply the change of variables formula.
To compute the Jacobian, we note that if we write $\varphi=(\varphi_1,\varphi_2), x=(x_1, x_2),$ and $w=(w_1, w_2),$ then $(\varphi_1(w),\varphi_2(w))=(x_1+tw_1, x_2+tw_2),$ and \begin{align*}\frac{\partial \varphi_1}{\partial w_1}&=t,\\ \frac{\partial \varphi_1}{\partial w_2}&=0,\\ \frac{\partial \varphi_2}{\partial w_1}&=t,\\ \frac{\partial \varphi_2}{\partial w_2}&=0, \end{align*} and we get the Jacobian $$D\varphi(w)=\begin{pmatrix} \frac{\partial \varphi_1}{\partial w_1}& \frac{\partial \varphi_1}{\partial w_2}\\ \frac{\partial \varphi_2}{\partial w_1} & \frac{\partial \varphi_2}{\partial w_2} \end{pmatrix}=\begin{pmatrix} t & 0 \\ 0 & t \end{pmatrix},$$ which has determinant $t^2.$