Get the answer to "Evaluate the following integral in cylindrical coordinates triple..."

iohanetc

Answered question

2021-05-12

Evaluate the following integral in cylindrical coordinates triple integral

\int_{x = - 1}^{1} \int_{y = 0}^{\sqrt{1 - x^{2}}} \int_{z = 0}^{2} (\frac{1}{1 + x^{2} + y^{2}}) d z d x d y

Elberte

Skilled2021-05-13Added 95 answers

Step 1
Evaluate integral by changing into cylindrical coordinates.
The formula for triple integration in cylindrical coordinates is

\int \int_{E} \int f (x, y, z) d V = \int_{α}^{β} \int_{h_{1} (θ)}^{h_{2} (θ)} \int_{u_{1} (r \cos θ, r \sin θ)}^{u_{2} (r \cos θ, r \sin θ)} f (r \cos θ, r \sin θ, z) r d z d r d θ

Step 2
Here, the region is

E = {(r, θ, z) | 0 \leq θ \leq 2 π, 0 \leq r \leq 1, 0 \leq z \leq 2}

\int_{x = - 1}^{1} \int_{y = 0}^{\sqrt{1 - x^{2}}} \int_{z = 0}^{2} (\frac{1}{1 + x^{2} + y^{2}}) d z d x d y

= \int_{θ = 0}^{2 π} \int_{r = 0}^{1} \int_{z = 0}^{2} (\frac{1}{1 + r^{2}}) r d z d r d θ

= \int_{θ = 0}^{2 π} \int_{r = 0}^{1} \frac{r}{1 + r^{2}} [z]_{0}^{2} d r d θ

= \int_{θ = 0}^{2 π} \int_{r = 0}^{1} \frac{2 r}{1 + r^{2}} d r d θ

= \int_{θ = 0}^{2 π} [\ln (1 + r^{2})]_{0}^{1} d θ

= \ln (2) \int_{θ = 0}^{2 π} d θ

= 2 π \ln (2)

Thus,

\int_{x = - 1}^{1} \int_{y = 0}^{\sqrt{1 - x^{2}}} \int_{z = 0}^{2} (\frac{1}{1 + x^{2} + y^{2}}) d z d x d y = 2 π \ln (2)

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