auto-update(nvim): 2025-03-06 02:01:44

documents/by-course/math-6a/course-notes/main.typ

@@ -1152,6 +1152,43 @@ $
   integral.double_(x^2 + y^2 <= 1) f(x,y) dif x dif y
 $
 
+#example[
+  Show both ways of setting up the integral of a function $f(x,y)$ over the region bounded by $y - x = 2$ and $y = x^2$.
+
+  Hopefully we can visualize what this looks like, now let's solve for the
+  intersection points.
+
+  $
+    (x,y) = (-1,1) "or" (2,4)
+  $
+
+  In AP Calculus we know we can compute this by subtracting the functions, but
+  let's consider it as a double integral instead.
+
+  We want the part of the plane that lies above the parabola, but below the
+  line. Let's think about this as a system of inequalities.
+
+  $
+    &y >= x^2 \
+    &y - x <= 2
+  $
+
+  First let's imagine we let $x$ be the outer integral. Solving for $y$,
+
+  $
+    x^2 <= y <= x + 2
+  $
+
+  Hence, we get the integral
+  $
+    integral^2_(x=-1) integral^(x+2)_(y=x^2)
+  $
+
+  If we put $y$ as the outer integral, we see that $y$ ranges from 0 to 4.
+  However we actually have a system of 3 inequalities. So we'll need to write a
+  sum of integrals and it's very annoying.
+]
+
 == Swapping the order of integration
 
 If our function is nice, then this is easy.
@@ -1166,3 +1203,36 @@ Otherwise, we want to swap the order of integration. We convert the limits of
 integration back into inequality/region format, getting a region $cal(R)$ like
 discussed in the previous section. Then evaluate that integral using standard
 methods.
+
+#example[
+  Evaluate the double integral
+  $
+    integral^2_(x=0) integral^1_(y=x / 2) e^(y^2) dif y dif x
+  $
+
+  To evaluate this, note that integrating $e^(y^2)$ is hard. So let's swap the
+  order. Convert to region format:
+  $
+    cal(R) = cases(0 <= x <= 2, x/2 <= y <= 1)
+  $
+  If we graph the region being integrated on an $x y$-plane, we see a
+  triangular area. Solving for $x$ in terms of $y$ gives us three conditions.
+  In addition to $0 <= x <= 2$ we need $x <= 2y$. Since $y <= 1$ we can ignore
+  $x <= 2$, and the region becomes
+  $
+    cal(R) = cases(0 <= y <= 1, 0 <= x <= 2y)
+  $
+  Converting back into double integral gives
+  $
+    integral_(y=0)^1 integral_(x=0)^(2y) e^(y^2) dif y dif x
+  $
+  The inner integral is with respect to $x$ but the integrand is independent of
+  $x$. So the inner integral is
+  $
+    integral^(2y)_(x=0) e^(y^2) dif x = 2y dot e^(y^2)
+  $
+  and evaluating
+  $
+    integral_(y=0)^1 (2y dot e^(y^2)) dif y dif x = lr(e^(y^2) |)^(y=1)_(y=0) = e - 1
+  $
+]