Как распаковать один файл из zip-архива в памяти?

using (ZipArchive archive = new ZipArchive(postedZipStream))
{
    foreach (ZipArchiveEntry entry in archive.Entries)
    {
         var stream = entry.Open();
         //Do awesome stream stuff!!
    }
}

string metaDataContents;
using (var zipStream = new FileStream(@"C:\OB10LinuxShare\TEST1\Temp" + "\\"+zipFileName+".zip", FileMode.Open))
using (var archive = new ZipArchive(zipStream, ZipArchiveMode.Read))
{
    foreach (var entry in archive.Entries)
    {
        if (entry.Name.EndsWith(".xml"))
        {
            FileInfo metadataFileInfo = new FileInfo(entry.Name);
            string metadataFileName = metadataFileInfo.Name.Replace(metadataFileInfo.Extension, String.Empty);
            if (String.Compare(zipFileName, metadataFileName, true) == 0)
            {
                using (var stream = entry.Open())
                using (var reader = new StreamReader(stream))
                {
                    metaDataContents = reader.ReadToEnd();
                    clientProcessLogWriter.WriteToLog(LogWriter.LogLevel.DEBUG, "metaDataContents : " + metaDataContents);
                }
            }
        }
    }
}