one more 'spacial'

diff --git a/examples/step-33/doc/results.dox b/examples/step-33/doc/results.dox
index 03b5889565bf2640cf38d15a139fcd54a7c0f630..74407af04e4c8ed6426561c4e1b753edbc8d62d5 100644 (file)
--- a/examples/step-33/doc/results.dox
+++ b/examples/step-33/doc/results.dox
@@ -266,7 +266,7 @@ reads
     , \mathbf{z}_i \right)_K  +
       \theta \mathbf{B}({\mathbf{w}^{k}_{n+1})(\mathbf{z}_i)_K +
       (1-\theta) \mathbf{B}({\mathbf{w}_{n}) (\mathbf{z}_i)_K $
-This means that we calculate the spacial residual twice at one Newton
+This means that we calculate the spatial residual twice at one Newton
 iteration step: once respect to the current solution ${\mathbf{w}^{k}_{n+1}$
 and once more respect to the last time step solution $\mathbf{w}_{n}$ which
 remains the same during all Newton iterations through one timestep.