# Thread:

1. ## polar coordinate

Hi please, can you help me to solve this integral using polar coordinate
Thank you

$<br /> \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}dxdy<br />$

2. $x^2+y^2=r^2$

Now you have $e^{-r^2}$ which is the same as doing polar since you have $x=rcos and y=rsin$ and squaring both removes the trig

3. $<br /> \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}dxdy=\int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2}r d \theta dr = 2\pi \int_0^{\infty} e^{-r^2}r dr = \pi [e^{-r^2}]_0^{\infty} = \pi<br /> <br />$
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9382538795471191, "perplexity_flag": "middle"}