Evaluate the improper integral.
-$$\int _ { - \infty } ^ { \infty } \int _ { - \infty } ^ { \infty } \frac { d x d y } { \left( x ^ { 2 } + 36 \right) \left( y ^ { 2 } + 9 \right) }$$
A) $\frac { \pi ^ { 2 } } { 36 }$
B) $\frac { \pi } { 36 }$
C) $\frac { \pi ^ { 2 } } { 18 }$
D) $\frac { \pi } { 18 }$

Question

Quizplus · Accepted Answer

We can evaluate this integral using the method of partial fractions.

First, we factor the denominator:

$$\frac { 1 } { \left( x ^ { 2 } + 36 \right) \left( y ^ { 2 } + 9 \right) } = \frac{A}{x^2+36} + \frac{B}{y^2+9}$$

We can find A and B by multiplying both sides by the denominator on the left and then setting x and y to appropriate values:

$$1 = A(y^2+9) + B(x^2+36)$$

Setting x=0, we get:

$$1 = A(y^2+9)$$

Setting y=0, we get:

$$1 = B(36)$$

So, we have:

$$A = \frac{1}{9}$$

and

$$B = \frac{1}{36}$$

Using these values, we can write:

$$\frac { 1 } { \left( x ^ { 2 } + 36 \right) \left( y ^ { 2 } + 9 \right) } = \frac{1}{9(x^2+36)} + \frac{1}{36(y^2+9)}$$

Now, we can evaluate the integral:

\begin{align*}
\int _ { - \infty } ^ { \infty } \int _ { - \infty } ^ { \infty } \frac { d x d y } { \left( x ^ { 2 } + 36 \right) \left( y ^ { 2 } + 9 \right) } &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{9(x^2+36)} + \frac{1}{36(y^2+9)} dxdy \
&= \int_{-\infty}^{\infty} \frac{1}{9(x^2+36)} dx \int_{-\infty}^{\infty} \frac{1}{36(y^2+9)} dy \
&= \frac{1}{9}\left(\frac{\pi}{6}\right) \left(\frac{\pi}{3}\right)\
&= \frac{\pi^2}{162}
\end{align*}

So the correct answer is C.

Evaluate the Improper Integral ∫−∞∞∫−∞∞dxdy(x2+36)(y2+9)\int _ { - \infty } ^ { \infty } \int _ { - \infty } ^ { \infty } \frac { d x d y } { \left( x ^ { 2 } + 36 \right) \left( y ^ { 2 } + 9 \right) }∫−∞∞​∫−∞∞​(x2+36)(y2+9)dxdy​

Evaluate the Improper Integral $\int _ { - \infty } ^ { \infty } \int _ { - \infty } ^ { \infty } \frac { d x d y } { \left( x ^ { 2 } + 36 \right) \left( y ^ { 2 } + 9 \right) }$