获取Web Service（POST方法）在C＃.NET中发送的XML

Question

我正在开发这个应用程序，我需要将XML文件发送到Web服务并从Web服务中读取它，并从Web服务应用业务逻辑，它将发送XML作为响应。

问题是我确实知道如何读取webservice发送的响应，但我不知道如何检索请求发送到webservice的XML，变量总是为空。如果有人可以帮助我，我会很高兴。

请求方法：

(...)
    using (var client = new HttpClient())
        {

            var XMLRequest = Util.BuildXML.CreateRequestXML();

            string url = string.Format(WEB_API_HOST + "/Application/MyMethod/");

            HttpRequestMessage httpRequest = new HttpRequestMessage()
            {
                RequestUri = new Uri(url, UriKind.Absolute),
                Method = HttpMethod.Post,
                Content = new StringContent(XMLRequest.ToString(), Encoding.UTF8, "text/xml")
            };

            var task = client.SendAsync(httpRequest).ContinueWith((taskwithmsg) =>
            {
                var response = taskwithmsg.Result;

                if (response.IsSuccessStatusCode)
                {
                    var xmlResponse = response.Content.ReadAsStringAsync().Result;
                    XmlDocument doc = new XmlDocument();
                    doc.LoadXml(xmlResponse);

                    var test = doc.SelectSingleNode("RESPONSE/...").InnerText;

                }
                else
                {
                    var contentTask = response.Content.ReadAsStringAsync().Result;

                    throw new Exception(contentTask);
                }
            });

            task.Wait();
        }
    (...)

Webservice（应该接收XML并读取它）

[AcceptVerbs("GET", "POST")]
    public HttpResponseMessage MyMethod(XmlDocument doc)
    {

        // Any doc.SelectSingleNode() wouldn't work, the doc variable is null.

        var response = new StringBuilder();

        response.Append("<?xml version=\"1.0\" standalone=\"yes\"?>");
        response.Append("<RESPONSE>");
        (...)
        response.Append("</RESPONSE>");

        var xmlResponse = new HttpResponseMessage()
        {
            Content = new StringContent(response.ToString(), Encoding.UTF8, "text/xml")
        };

        return xmlResponse;
    }

那么如何在“MyMethod”方法中阅读xml？

非常感谢！